AP calculus AB, 2 related problems

[BifSlamkovich]

Homework Statement

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(x²), find the derivative of h'(x)? Lower bound on the integral is 2-3x and upper bound is x².

Homework Equations

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 x² - the integral of f(t) evaluated at 2-3x. If g(x) is equal to this, then the derivative of it is 2x*f(x²) - 3*f(2-3x), since where taking derivative wrt x. Then finally, we plug in x² wherever there is an x to get h'(x) (=g'(x²)

 

[SammyS]

Homework Statement

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(x²), find the derivative of h'(x)? Lower bound on the integral is 2-3x and upper bound is x².

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 x² - the integral of f(t) evaluated at 2-3x. If g(x) is equal to this, then the derivative of it is 2x*f(x²) - 3*f(2-3x), since where taking derivative wrt x. Then finally, we plug in x² wherever there is an x to get h'(x) (=g'(x²)

First Problem: Yes, you apply the chain rule.

The chain rule gives $\displaystyle h'(x)=g'\left(\frac{2}{x}\right)\cdot\frac{d}{dx} \left(\frac{2}{x}\right)\,.$

Second problem:

Suppose $F(x)=\int\ f(x)\ dx$, then $F\,'(x) = f(x)\,.$

Your integral is then given by: $\displaystyle \int_{x^2}^{2-3x}\ f(t)\ dt = F(x^2)-F(2-3x)$

Take the derivative of that. Don't forget the chain rule & that $F\,'(x) = f(x)\,.$

 

[BifSlamkovich]

Thanks for helping me to understand the first problem.

For the second problem, I am wondering do we plug in x² for x before or after we differentiate, since h(x) = g(x²)? What is the reason for this?

 

[BifSlamkovich]

Thanks for helping me to understand the first problem.

For the second problem, I am wondering do we plug in x² for x before or after we differentiate, since h(x) = g(x²)? What is the reason for this?

Never mind, I figured it out. Thanks anyway for all the help!