Derivative of the square root of the function f(x squared)

[Strand9202]

Homework Statement

I was asked to find the derivative of the sqrt (f(x^2)).

Relevant Equations

Chain Rule

I started out by rewriting the function as (f(x^2))^(1/2). I then did chain rule to get 1/2(f(x^2))^(-1/2) *(f'(x^2).

- I think I need to go further because it is an x^2 in the function, but not sure.

 

[PeroK]

- I think I need to go further because it is an x^2 in the function, but not sure.

Yes, you do. Effectively you have $g(f(h(x)))$, where $g(x) = x^{1/2}$ and $h(x) = x^2$.

 

[Strand9202]

Yes, you do. Effectively you have $g(f(h(x)))$, where $g(x) = x^{1/2}$ and $h(x) = x^2$.

Would I be wise and just restarting all over again, or should I just pick up from where I left off.

 

[PeroK]

Would I be wise and just restarting all over again, or should I just pick up from where I left off.

That's up to you.

 

[Strand9202]

Homework Statement:: I was asked to find the derivative of the sqrt (f(x^2)).
Relevant Equations:: Chain Rule

I started out by rewriting the function as (f(x^2))^(1/2). I then did chain rule to get 1/2(f(x^2))^(-1/2) *(f'(x^2).

- I think I need to go further because it is an x^2 in the function, but not sure.

I have tried and gone further. I got to the final answer of
1/2(f(x^2))^(-1/2) *f'(x^2)*2x

Is that my final answer now?

 

[Strand9202]

That's up to you.

I have tried and gone further. I got to the final answer of
1/2(f(x^2))^(-1/2) *f'(x^2)*2x

 

[PeroK]

I have tried and gone further. I got to the final answer of
1/2(f(x^2))^(-1/2) *f'(x^2)*2x

Here's an idea. Try with $f(x) =$ some function and see whether you get the right answer.

 

[Strand9202]

Here's an idea. Try with $f(x) =$ some function and see whether you get the right answer.

I did that. I let f(x)=sin. That way my overall function reads square root (sin(x^2)). I got that my final answer was 1/2(sin(x^2))^(-1/2) * -cos(x^2)*2x. Which does fit with my rule. Therefore that shows my answer is correct.

 

[PeroK]

I did that. I let f(x)=sin. That way my overall function reads square root (sin(x^2)). I got that my final answer was 1/2(sin(x^2))^(-1/2) * -cos(x^2)*2x. Which does fit with my rule. Therefore that shows my answer is correct.

You could also have tried $f(x) = \sin^2 x$, where $f'(x) = 2 \sin x \cos x$. Then we have: $$k(x) = \sqrt{f(x^2)} = \sin x^2$$ and $$k'(x) = 2x \cos x^2$$ Then, using your formula we have: $$k'(x) = \frac 1 2 \frac 1 {\sin x^2} (2\sin x^2 \cos x^2) (2x) = 2x \cos x^2$$ PS Check the derivative of $\sin x$.