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.
The derivative of the square root of the function f(x squared) is equal to 1/2 times the reciprocal of the square root of f(x squared) multiplied by the derivative of f(x squared).
To find the derivative of the square root of a function, you can use the power rule and chain rule. First, rewrite the function as f(x)^(1/2) and then apply the power rule to get 1/2*f(x)^(-1/2). Next, use the chain rule to multiply by the derivative of the function inside the square root.
Yes, the derivative of the square root of a function can be negative. This will occur when the function inside the square root is decreasing, causing the derivative to be negative.
The derivative of the square root of a function is the slope of the tangent line at a specific point on the function, while the square root of the derivative of a function is a function itself. The two may have different values at different points on the original function.
Yes, the derivative of the square root of a function can be undefined. This will occur when the function inside the square root is equal to zero, as the reciprocal of zero is undefined.