Yes, you do. Effectively you have ##g(f(h(x)))##, where ##g(x) = x^{1/2}## and ##h(x) = x^2##.Strand9202 said:- I think I need to go further because it is an x^2 in the function, but not sure.
Would I be wise and just restarting all over again, or should I just pick up from where I left off.PeroK said:Yes, you do. Effectively you have ##g(f(h(x)))##, where ##g(x) = x^{1/2}## and ##h(x) = x^2##.
That's up to you.Strand9202 said:Would I be wise and just restarting all over again, or should I just pick up from where I left off.
Strand9202 said: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 ofPeroK said:That's up to you.
Here's an idea. Try with ##f(x) = ## some function and see whether you get the right answer.Strand9202 said:I have tried and gone further. I got to the final answer of
1/2(f(x^2))^(-1/2) *f'(x^2)*2x
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 said:Here's an idea. Try with ##f(x) = ## some function and see whether you get the right answer.
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##.Strand9202 said: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.