The discussion revolves around finding the derivative of the function defined as the square root of another function, specifically f(x^2). Participants are exploring the application of the chain rule in this context.
Some participants have provided guidance on how to approach the problem, suggesting that testing with specific functions could help verify the correctness of the derivative obtained. Multiple interpretations of the problem are being explored, with no explicit consensus reached.
Participants mention the relevance of the chain rule and have attempted to apply it to different functions to check their results. There is an ongoing discussion about the correctness of their derived expressions.
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.