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Derivative of complex equation

  1. Dec 6, 2015 #1
    hello!
    1) what is the process to get the derivative of an equation that requires you to do first the chain rule and then the product/quotient rule, eg. sin(x^2(x+1))?
    2) what is the process to get the derivative of an equation that requires you to do first the product/quotient rule and then the chain rule, eg. sin(e^x+1)(cos(x^2+1))?
    thanks!
     
  2. jcsd
  3. Dec 6, 2015 #2

    FactChecker

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    Step by step. Just do the first step first leaving d/dx symbols where they belong. Then see if more can be done to the resulting parts. There are too many combinations to try to memorize formulas for them all. Baby steps gets the job done.

    I assume you mean (x^2)*(x+1) and not x^(2*(x+1)). Parentheses are your friends. Use them wherever it might otherwise be ambiguous.

    d/dx( sin((x^2)(x+1)) ) = cos((x^2)(x+1)) * d/dx ((x^2)(x+1))
    = cos((x^2)(x+1)) * [ d/dx x^2 ]*(x+1) + (x^2)[d/dx(x+1)])
    = cos((x^2)(x+1)) * [ (2*x(x+1) + x^2) ]
     
  4. Dec 6, 2015 #3

    Simon Bridge

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    1. You just described it ... you do the chain rule first then the product/quotient rule.
    2. Same here... you just described the process. It is the process of using the product rule followed by the pricess of doing the chain rule.

    The trick is to do it in strps.

    Eg 1.

    ##\frac{d}{dx}\left ( \sin\left ( x^2 (x+1) \right) \right)## ...put u=x^2 (x+1) and apply the chain rule.

    ##= \cos u \frac{du}{dx}## now evaluate du/dx using the product rule.

    ##\frac{du}{dx} = 2x(x+1) + x^2##

    ... now substitute back.
     
  5. Dec 6, 2015 #4

    Mark44

    Staff: Mentor

    stirrups? As in, astride a horse? :oldbiggrin:
     
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