Determine the second derivative of a function

[kent davidge]

Homework Statement

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a function two times differentiable and $g: \mathbb{R} \rightarrow \mathbb{R}$ given by $g(x) = f(x + 2 \cos(3x))$.

(a) Determine g''(x).

(b) If f'(2) = 1 and f''(2) = 8, compute g''(0).

Homework Equations

I'm not aware of any.

The Attempt at a Solution

[/B]
I was thinking about a Taylor expansion of $f$ around $x$. Is it allowed? Anyway, it was getting very complicated because of the derivatives getting higher and higher (and we have information that $f$ is two times differentiable, no guarantee that it's more than that).

So
for (a) I simply answered that g''(x) = f''(x + 2 cos(3x)). I'm not sure that's enough.
for (b) I noticed that g(0) = f(2) and I asserted that g''(0) = f''(2) = 8. But I do not think that's right because the problem even give f'(2). It would not do it without a reason.

 

[tnich]

Homework Statement

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a function two times differentiable and $g: \mathbb{R} \rightarrow \mathbb{R}$ given by $g(x) = f(x + 2 \cos(3x))$.

(a) Determine g''(x).

(b) If f'(2) = 1 and f''(2) = 8, compute g''(0).

Homework Equations

I'm not aware of any.

The Attempt at a Solution

[/B]
I was thinking about a Taylor expansion of $f$ around $x$. Is it allowed? Anyway, it was getting very complicated because of the derivatives getting higher and higher (and we have information that $f$ is two times differentiable, no guarantee that it's more than that).

So
for (a) I simply answered that g''(x) = f''(x + 2 cos(3x)). I'm not sure that's enough.
for (b) I noticed that g(0) = f(2) and I asserted that g''(0) = f''(2) = 8. But I do not think that's right because the problem even give f'(2). It would not do it without a reason.

You need to use the chain rule.

 

[kent davidge]

You need to use the chain rule

Should I use it for question (a) or (b)?

 

[tnich]

Should I use it for question (a) or (b)?

You should use if for a), and then substitute values for x and the derivatives of f for b).

 

[Mark44]

Thread moved.
@kent davidge, please post questions about derivatives (and integrals) in the Calculus & Beyond section, not in the Precalc section.

 

[fishspawned]

consider that one way of thinking about derivatives with the chain rule is that you have a function INSIDE another function. so start with the idea that a chain rule assumes you have a function of the form f(g(x)). what does the derivative look like, according to chain rule?

 

[kent davidge]

Thanks for the answers. Using the chain rule I get:

(a) $g''(x) = f''(h(x))[h'(x)]^2 + f'(h(x))h''(x)$

(b) $g''(0) = f''(h(0))[h'(0)]^2 + f'(h(0))h''(0) = 1 \times f''(2) - 18 \times f'(2) = -143.$

Correct?

 

[PeroK]

Thanks for the answers. Using the chain rule I get:

(a) $g''(x) = f''(h(x))[h'(x)]^2 + f'(h(x))h''(x)$

That's good. Although they probably expect a specific answer using the given $h(x)$

(b) $g''(0) = f''(h(0))[h'(0)]^2 + f'(h(0))h''(0) = 1 \times f''(2) - 18 \times f'(2) = -143.$

Correct?

Not so good.

 

[kent davidge]

Not so good.

It used the wrong values, is it? The correct result is $-10$.

 

[tnich]

It used the wrong values, is it? The correct result is $-10$.

You used the right values, they were just in the wrong places.