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2.7 chain rule short cuts

The document discusses various chain rules for derivatives, including: - The power chain rule: [up]' = pup−1(u)' - Trigonometric chain rules: [sin(u)]' = cos(u)(u)', [cos(u)]' = −sin(u)(u)' - Examples are provided to demonstrate applying the chain rules to find derivatives of more complex functions like y = sin(x3) and y = sin3(x). Repeated application of the appropriate chain rule at each step is often required.

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Chain Rule Short Cuts
Chain Rule Short Cuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules.
Chain Rule Short Cuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' b. [sin(u)]' = cos(u)*(u)' c. [cos(u)]' = –sin(u)*(u)' (The Power Chain Rule) (The Sine Chain Rule) (The Cosine Chain Rule)
Chain Rule Short Cuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x.
Chain Rule Short Cuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2.
Chain Rule Short Cuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)'
Chain Rule Short Cuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)' to get [(x3+1)2]'
Chain Rule Short Cuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)' to get [(x3+1)2]' = 2(x3+1)2–1
Chain Rule Short Cuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)' to get [(x3+1)2]' = 2(x3+1)2–1(x3+1)' u'
Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' b. [sin(u)]' = cos(u)*(u)' c. [cos(u)]' = –sin(u)*(u)' where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)' to get = 6x2 (x3+1) Chain Rule Short Cuts = 2(x3+1)3x2 (The Power Chain Rule) (The Sine Chain Rule) (The Cosine Chain Rule) [(x3+1)2]' = 2(x3+1)2–1(x3+1)' u'
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2]
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)'
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' =
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]'
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2 = –sin[(x3+1)2] [2(x3+1)1](x3+1)'
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2 = –sin[(x3+1)2] [2(x3+1)1](x3+1)' = –sin[(x3+1)2] [2(x3+1)]3x2
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2 = –sin[(x3+1)2] [2(x3+1)1](x3+1)' = –sin[(x3+1)2] [2(x3+1)]3x2 We list the frequently used Chain Rules below.
Chain Rule Short Cuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2 = –sin[(x3+1)2] [2(x3+1)1](x3+1)' = –sin[(x3+1)2] [2(x3+1)]3x2 We list the frequently used Chain Rules below. The list includes the log and exponential functions for the sole purpose of practicing symbolic algebraic manipulation. The verifications will be given later.
Chain Rule Short Cuts (Chain Rules Short Cuts) (The Power Chain Rule) [up]' = pup–1(u)' (The Trig. Chain Rules) [sin(u)] ' = sin'(u) = cos(u)(u)' [cos(u)] ' = cos'(u) = –sin(u)(u)' [tan(u)] ' = tan'(u) = sec2(u)(u)' [cot(u)] ' = cot'(u) = –csc2(u)(u)' [sec (u)] ' = sec'(u) = sec(u)tan(u)(u)' [csc (u)] ' = csc'(u) = –csc(u)cot(u)(u)' (The Derivatives of Log and Exponential Functions) and [eu]' = eu(u)' and [ln(u)]' = (u)' u [ex]' = ex [ln(x)]' = x 1
dup dx d sin(u) dx Chain Rule Short Cuts Below are the d/dx versions of the same rules. = pup–1du dx = cos(u) du dx d cos(u) dx = –sin(u) du dx d tan(u) dx = sec2(u)du dx d cot(u) dx = –csc2(u) du dx d sec(u) dx = sec(u)tan(u)du dx d csc(u) dx = –csc(u)cot(u)du dx deu dx = eu du dx d ln(u) dx = du 1 u dx
Chain Rule Short Cuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function.
Chain Rule Short Cuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression.
Chain Rule Short Cuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3)
Chain Rule Short Cuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3).
Chain Rule Short Cuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3). The last operation applied is sin(u) so we view sin(x3) as sin(u) with u = x3.
Chain Rule Short Cuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3). The last operation applied is sin(u) so we view sin(x3) as sin(u) with u = x3. Hence dx = cos(x3) d sin(x3)
Chain Rule Short Cuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3). The last operation applied is sin(u) so we view sin(x3) as sin(u) with u = x3. dx = d x3 dx Hence d sin(x cos(x3) 3)
Chain Rule Short Cuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3). The last operation applied is sin(u) so we view sin(x3) as sin(u) with u = x3. dx = d x3 dx Hence d sin(x cos(x3) 3) = 3x2cos(x3)
Chain Rule Short Cuts b. y = sin3(x)
Chain Rule Short Cuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x).
Chain Rule Short Cuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]'
Chain Rule Short Cuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]'
Chain Rule Short Cuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]' = 3sin2(x)[sin(x)]'
Chain Rule Short Cuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]' = 3sin2(x)[sin(x)]' = 3sin2(x)cos(x)
Chain Rule Short Cuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]' = 3sin2(x)[sin(x)]' = 3sin2(x)cos(x) c. y = ln(5e3x – 2)
Chain Rule Short Cuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]' = 3sin2(x)[sin(x)]' = 3sin2(x)cos(x) c. y = ln(5e3x – 2) The last operation applied is the log–function so we use the Log Chain Rule with u = (5e3x – 2).
Chain Rule Short Cuts d ln(5e3x – 2) dx d ln(u) dx = du 1 u dx deu dx = eu du dx
Chain Rule Short Cuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx d ln(u) dx = du 1 u dx deu dx = eu du dx
Chain Rule Short Cuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx deu dx = eu du dx
Chain Rule Short Cuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx the new u deu dx = eu du dx
Chain Rule Short Cuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx the new u deu dx = eu du dx = (5e3x – 2) d 3x dx 5 e3x
Chain Rule Short Cuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx the new u deu dx = eu du dx = (5e3x – 2) d 3x dx 5 e3x = 15 e3x (5e3x – 2)
Chain Rule Short Cuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx the new u deu dx = eu du dx = 5 e3x (5e3x – 2) d 3x dx = 15 e3x (5e3x – 2) If the last operation applied in the expression were the algebraic operations of multiplication or division by a non–constant, then the Product and Quotient Rules should be applied respectively.
Chain Rule Short Cuts d. y = ex2+1 ln(x)
Chain Rule Short Cuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first.
Chain Rule Short Cuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' gf ' – fg' g2 f g ( )' =
Chain Rule Short Cuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = x2+1 ln(x)[e ]' – ln2(x) gf ' – fg' g2 f g ( )' = ex2+1[ln(x)]'
Chain Rule Short Cuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = x2+1 ln(x)[e ]' – ex2+1[ln(x)]' ln2(x) 1/x gf ' – fg' g2 f g ( )' =
Chain Rule Short Cuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = Use the Exponential Chain Rule x2+1 ln(x)[e ]' – ex2+1[ln(x)]' ln2(x) 1/x gf ' – fg' g2 f g ( )' =
Chain Rule Short Cuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = x2+1 ln(x)[e ]' – ex2+1[ln(x)]' ln2(x) = x2+1 ln(x)[e ][x2+1]' – ln2(x) ex2+1/x Use the Exponential Chain Rule 1/x u' = gf ' – fg' g2 f g ( )' =
Chain Rule Short Cuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = x2+1 ln(x)[e ]' – ex2+1[ln(x)]' ln2(x) = x2+1 ln(x)[e ][x2+1]' – ln2(x) ex2+1/x Use the Exponential Chain Rule 1/x u' = gf ' – fg' x2+1 e [2x2ln(x) – 1] xln2(x) g2 f g ( )' =

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2.7 chain rule short cuts

  • 1.
  • 2.
    Chain Rule ShortCuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules.
  • 3.
    Chain Rule ShortCuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' b. [sin(u)]' = cos(u)*(u)' c. [cos(u)]' = –sin(u)*(u)' (The Power Chain Rule) (The Sine Chain Rule) (The Cosine Chain Rule)
  • 4.
    Chain Rule ShortCuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x.
  • 5.
    Chain Rule ShortCuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2.
  • 6.
    Chain Rule ShortCuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)'
  • 7.
    Chain Rule ShortCuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)' to get [(x3+1)2]'
  • 8.
    Chain Rule ShortCuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)' to get [(x3+1)2]' = 2(x3+1)2–1
  • 9.
    Chain Rule ShortCuts Set u = u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' (The Power Chain Rule) b. [sin(u)]' = cos(u)*(u)' (The Sine Chain Rule) c. [cos(u)]' = –sin(u)*(u)' (The Cosine Chain Rule) where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)' to get [(x3+1)2]' = 2(x3+1)2–1(x3+1)' u'
  • 10.
    Set u =u(x) and y = up where p is a number, or y = cos(u), or y = sin(u), we get the following rules. (Specific Chain Rules) a. [up]' = pup–1(u)' b. [sin(u)]' = cos(u)*(u)' c. [cos(u)]' = –sin(u)*(u)' where ( u )' is the back-derivative with respect to x. Example A. a. Find the derivative of (x3+1)2. Set the base u = (x3+1), p = 2 and use the Power Rule [up]' = pup–1(u)' to get = 6x2 (x3+1) Chain Rule Short Cuts = 2(x3+1)3x2 (The Power Chain Rule) (The Sine Chain Rule) (The Cosine Chain Rule) [(x3+1)2]' = 2(x3+1)2–1(x3+1)' u'
  • 11.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2]
  • 12.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)'
  • 13.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' =
  • 14.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]'
  • 15.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2
  • 16.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2 = –sin[(x3+1)2] [2(x3+1)1](x3+1)'
  • 17.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2 = –sin[(x3+1)2] [2(x3+1)1](x3+1)' = –sin[(x3+1)2] [2(x3+1)]3x2
  • 18.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2 = –sin[(x3+1)2] [2(x3+1)1](x3+1)' = –sin[(x3+1)2] [2(x3+1)]3x2 We list the frequently used Chain Rules below.
  • 19.
    Chain Rule ShortCuts b. Find the derivative of cos[(x3+1)2] Set the cosine input (x3+1)2 = u, and use the Cosine Rule cos(u)' = –sin(u)(u)' to get (cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]' the new “u” for u2 = –sin[(x3+1)2] [2(x3+1)1](x3+1)' = –sin[(x3+1)2] [2(x3+1)]3x2 We list the frequently used Chain Rules below. The list includes the log and exponential functions for the sole purpose of practicing symbolic algebraic manipulation. The verifications will be given later.
  • 20.
    Chain Rule ShortCuts (Chain Rules Short Cuts) (The Power Chain Rule) [up]' = pup–1(u)' (The Trig. Chain Rules) [sin(u)] ' = sin'(u) = cos(u)(u)' [cos(u)] ' = cos'(u) = –sin(u)(u)' [tan(u)] ' = tan'(u) = sec2(u)(u)' [cot(u)] ' = cot'(u) = –csc2(u)(u)' [sec (u)] ' = sec'(u) = sec(u)tan(u)(u)' [csc (u)] ' = csc'(u) = –csc(u)cot(u)(u)' (The Derivatives of Log and Exponential Functions) and [eu]' = eu(u)' and [ln(u)]' = (u)' u [ex]' = ex [ln(x)]' = x 1
  • 21.
    dup dx dsin(u) dx Chain Rule Short Cuts Below are the d/dx versions of the same rules. = pup–1du dx = cos(u) du dx d cos(u) dx = –sin(u) du dx d tan(u) dx = sec2(u)du dx d cot(u) dx = –csc2(u) du dx d sec(u) dx = sec(u)tan(u)du dx d csc(u) dx = –csc(u)cot(u)du dx deu dx = eu du dx d ln(u) dx = du 1 u dx
  • 22.
    Chain Rule ShortCuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function.
  • 23.
    Chain Rule ShortCuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression.
  • 24.
    Chain Rule ShortCuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3)
  • 25.
    Chain Rule ShortCuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3).
  • 26.
    Chain Rule ShortCuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3). The last operation applied is sin(u) so we view sin(x3) as sin(u) with u = x3.
  • 27.
    Chain Rule ShortCuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3). The last operation applied is sin(u) so we view sin(x3) as sin(u) with u = x3. Hence dx = cos(x3) d sin(x3)
  • 28.
    Chain Rule ShortCuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3). The last operation applied is sin(u) so we view sin(x3) as sin(u) with u = x3. dx = d x3 dx Hence d sin(x cos(x3) 3)
  • 29.
    Chain Rule ShortCuts Usually repeated applications of the above rules are required to compute the derivative of an arbitrary elementary function. The particular chain rule to be used at a particular step corresponds to the last function applied in the expression. Example B. Find the following derivatives. a. y = sin(x3) The formula takes the input x→x3→sin(x3). The last operation applied is sin(u) so we view sin(x3) as sin(u) with u = x3. dx = d x3 dx Hence d sin(x cos(x3) 3) = 3x2cos(x3)
  • 30.
    Chain Rule ShortCuts b. y = sin3(x)
  • 31.
    Chain Rule ShortCuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x).
  • 32.
    Chain Rule ShortCuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]'
  • 33.
    Chain Rule ShortCuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]'
  • 34.
    Chain Rule ShortCuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]' = 3sin2(x)[sin(x)]'
  • 35.
    Chain Rule ShortCuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]' = 3sin2(x)[sin(x)]' = 3sin2(x)cos(x)
  • 36.
    Chain Rule ShortCuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]' = 3sin2(x)[sin(x)]' = 3sin2(x)cos(x) c. y = ln(5e3x – 2)
  • 37.
    Chain Rule ShortCuts b. y = sin3(x) The formula takes the input x→sin(x)→[sin(x)]3 so the last operation applied is u3 where u = sin(x). Using the Power Rule with u3 we have [sin3(x)]' = [u3]' = 3u2[u]' = 3sin2(x)[sin(x)]' = 3sin2(x)cos(x) c. y = ln(5e3x – 2) The last operation applied is the log–function so we use the Log Chain Rule with u = (5e3x – 2).
  • 38.
    Chain Rule ShortCuts d ln(5e3x – 2) dx d ln(u) dx = du 1 u dx deu dx = eu du dx
  • 39.
    Chain Rule ShortCuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx d ln(u) dx = du 1 u dx deu dx = eu du dx
  • 40.
    Chain Rule ShortCuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx deu dx = eu du dx
  • 41.
    Chain Rule ShortCuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx the new u deu dx = eu du dx
  • 42.
    Chain Rule ShortCuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx the new u deu dx = eu du dx = (5e3x – 2) d 3x dx 5 e3x
  • 43.
    Chain Rule ShortCuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx the new u deu dx = eu du dx = (5e3x – 2) d 3x dx 5 e3x = 15 e3x (5e3x – 2)
  • 44.
    Chain Rule ShortCuts d ln(5e3x – 2) dx = 1 d (5e3x – 2) (5e3x – 2) dx = 1 5 d (e3x) (5e3x – 2) dx d ln(u) dx = du 1 u dx the new u deu dx = eu du dx = 5 e3x (5e3x – 2) d 3x dx = 15 e3x (5e3x – 2) If the last operation applied in the expression were the algebraic operations of multiplication or division by a non–constant, then the Product and Quotient Rules should be applied respectively.
  • 45.
    Chain Rule ShortCuts d. y = ex2+1 ln(x)
  • 46.
    Chain Rule ShortCuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first.
  • 47.
    Chain Rule ShortCuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' gf ' – fg' g2 f g ( )' =
  • 48.
    Chain Rule ShortCuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = x2+1 ln(x)[e ]' – ln2(x) gf ' – fg' g2 f g ( )' = ex2+1[ln(x)]'
  • 49.
    Chain Rule ShortCuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = x2+1 ln(x)[e ]' – ex2+1[ln(x)]' ln2(x) 1/x gf ' – fg' g2 f g ( )' =
  • 50.
    Chain Rule ShortCuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = Use the Exponential Chain Rule x2+1 ln(x)[e ]' – ex2+1[ln(x)]' ln2(x) 1/x gf ' – fg' g2 f g ( )' =
  • 51.
    Chain Rule ShortCuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = x2+1 ln(x)[e ]' – ex2+1[ln(x)]' ln2(x) = x2+1 ln(x)[e ][x2+1]' – ln2(x) ex2+1/x Use the Exponential Chain Rule 1/x u' = gf ' – fg' g2 f g ( )' =
  • 52.
    Chain Rule ShortCuts d. y = ex2+1 ln(x) The last operation executed in the formula is division so we use the Quotient Rule first. ex2+1 ln(x) [ ]' = x2+1 ln(x)[e ]' – ex2+1[ln(x)]' ln2(x) = x2+1 ln(x)[e ][x2+1]' – ln2(x) ex2+1/x Use the Exponential Chain Rule 1/x u' = gf ' – fg' x2+1 e [2x2ln(x) – 1] xln2(x) g2 f g ( )' =