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Topic: [ap-calculus] Need help plaease on 2008 BC MC # 17
Replies: 1   Last Post: Sep 20, 2012 4:48 PM

 Richard Sisley Posts: 4,189 Registered: 12/6/04
Re: [ap-calculus] Need help plaease on 2008 BC MC # 17
Posted: Sep 20, 2012 4:48 PM

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On Sep 20, 2012, at 1:18 PM, Qayumi, Enayat wrote:

> ---------------------------------------------------------------------------------------------
> If h is a differentiable function and f(x) = h (x^2 - 3)
>
> Why is the f'(x) = 2x h'(x^2-3) which will give the correct answer once you plug in x = 2
>
> And Not
>
> f'(x) = h' (x^2 - 3) + h (2x)

There is a difference between f(x)=h(x)*(x^2-3) and f(x)=h(x^2-3). In the first case f is given as the product of a function named "h" and a second function (lets call it "g") with g(x) = x^2 - 3. so f(x) = h(x) * g(x). In the second case f is given as the composite of h following g. So f(x) = h(g(x)). Try the Chain Rule instead of the Product Rule.

Sincerely,

Richard Sisley
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