Find the derivative of
f(x) =cos^-1(x^2)
Ok so I'm having a little confusion with the chain rule and I'm not sure if my answer is right...
will the answer be... 2x(-cos^-2(x^2)(-sin(x^2)?
recall that the derivative of cos^-1 (x) or arccos (x) is
-1/[sqrt(1-x^2)] * dx
thus the derivative of cos^-1 (x^2)
-(2x)/[sqrt(1 - x^4)]
note that the 2x is the dx that came from the derivatice of the term inside the cos^-1, which is x^2.
hope this helps~ :)
In google type:
calc101
When you see list of results click on first link.
When page be open click option derivatives.
When that page be open in rectangle type your function and click option DO IT.
You will see solution step by step.
If cos^-1(x^2) mean 1/cos(x^2)
type cos[x^2]^(-1)
If cos^-1(x^2) mean arccos(x^2)
type arccos[x^2]
To find the derivative of f(x) = cos^(-1)(x^2), we can use the chain rule. The chain rule states that if we have a composition of functions, f(g(x)), then the derivative is given by f'(g(x)) * g'(x).
Let's start by determining the derivative of the outer function, f(x) = cos^(-1)(x). The derivative of arccos(x) is equal to -1 / sqrt(1 - x^2).
Now, let's determine the derivative of the inner function, g(x) = x^2. The derivative of x^2 with respect to x is 2x.
Multiplying these derivatives together using the chain rule, we have:
f'(x) = (-1 / sqrt(1 - x^2)) * (2x)
Combining this expression, we get:
f'(x) = -2x / sqrt(1 - x^2)
Therefore, the correct answer is f'(x) = -2x / sqrt(1 - x^2).