1. ## calculate : f"'(x)

if:
 f(sinx)=(sinx)² calculate : f"(x)

2. ## Re: calculate : f"'(x)

The function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by

$f(x)=x^2$

is twice differentiable and satisfies the condition

$f(\sin (x))=(\sin x)^2$ for all $x \in \mathbb{R}$

3. ## Re: calculate : f"'(x)

Originally Posted by dhiab
if:
 f(sinx)=(sinx)² calculate : f"(x)
Use the chain rule:

Let y = sin(x). Then
$\displaystyle f(y) = y^2$

and
$\displaystyle \dfrac{df}{dx} = \dfrac{df}{dy} \dfrac{dy}{dx} = 2y \cdot cos(x) = 2~sin(x)~cos(x) = sin(2x)$

Can you finish?

-Dan

4. ## Re: calculate : f"'(x)

Originally Posted by topsquark
Use the chain rule:

Let y = sin(x). Then
$\displaystyle f(y) = y^2$
No! That would be correct if $\displaystyle f(x)= sin^2(x)$ but the original post said $\displaystyle f(sin(x))= sin^2(x)$. As Idea said, that implies that $\displaystyle f(x)= x^2$ so $\displaystyle f'(x)= 2x$ and $\displaystyle f''(x)= 2$.

and
$\displaystyle \dfrac{df}{dx} = \dfrac{df}{dy} \dfrac{dy}{dx} = 2y \cdot cos(x) = 2~sin(x)~cos(x) = sin(2x)$

Can you finish?

-Dan

5. ## Re: calculate : f"'(x)

Originally Posted by frick
No! That would be correct if $\displaystyle f(x)= sin^2(x)$ but the original post said $\displaystyle f(sin(x))= sin^2(x)$. As Idea said, that implies that $\displaystyle f(x)= x^2$ so $\displaystyle f'(x)= 2x$ and $\displaystyle f''(x)= 2$.
Yes, like I said $\displaystyle f(sin(x)) = sin^2(x)$. Yeah. That's what I wrote.

Thanks for the catch!

-Dan