(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

First problem:

Data in the form of a table is given for all integers x in [-2,3], f(x), g(x), f'(x), and g'(x)

Given that h(x) = g(2/x), find h'(2/3)

Second problem:

Given g(x) = ∫_{2-3x}^{x2}f(t) dt, and h(x) = g(x^{2}), find the derivative of h'(x)? Lower bound on the integral is 2-3x and upper bound is x^{2}.

2. Relevant equations

3. The attempt at a solution

First problem:

Do we or do we not apply the chain rule? Since h(x) = g(2/x), then does h'(x) = g'(2/x)?

Second problem:

We can simplify g(x) to the integral f(t) evaluated at x^{2}- the integral of f(t) evaluated at 2-3x. If g(x) is equal to this, then the derivative of it is 2x*f(x^{2}) - 3*f(2-3x), since where taking derivative wrt x. Then finally, we plug in x^{2}wherever there is an x to get h'(x) (=g'(x^{2})

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# Homework Help: AP calculus AB, 2 related problems

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