Find the derivative of the given function.f(x)=tan−1(x2)f(x)=\tan^{-1}(x^2)f(x)=tan−1(x2)

2 x 1 + x 4 \frac{2x}{1+x^4} 1 + x 4 2 x

Find the derivative of the given function. f ( x ) = tan − 1 ( x 2 ) f(x)=\tan^{-1}(x^2) f ( x ) = tan − 1 ( x 2 )

Here, we wanna find the derivative of the function f of x is equal to the inverse tangent of x squared. So in order to find this derivative f prime of x, we're gonna start with that outermost function and work our way inside using the chain rule. So the derivative of the inverse tangent is going to be 1 over 1 plus whatever we're taking the inverse tangent of, so in this case, x squared, and that needs to get squared. Now using the chain rule here, we then need to multiply by this derivative of x squared, which is going to be 2 x. Now we can clean this up a little bit, condensing this fraction. This is 2 x over 1 plus x squared squared is going to be x to the power of 4, and this is our derivative here, f prime of x. Let us know if you have any questions. I'll see you in the next one.

Find the derivative of the given function. f ( x ) = tan ⁡ − 1 ( x 2 ) f(x)=\tan^{-1}(x^2) f ( x ) = tan − 1 ( x 2 )

2 x 1 + x 4 \frac{2x}{1+x^4} 1 + x 4 2 x

2 x 1 + x 2 \frac{2x}{1+x^2} 1 + x 2 2 x

1 1 + x 4 \frac{1}{1+x^4} 1 + x 4 1

sec ⁡ 2 ( x 2 ) \sec^2\left(x^2\right) sec 2 ( x 2 )

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

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Multiple Choice

Find the derivative of the given function.
$f(x)=\tan^{-1}(x^2)$

$\frac{2x}{1+x^4}$

$\frac{2x}{1+x^2}$

$\frac{1}{1+x^4}$

$\sec^2\left(x^2\right)$

Verified step by step guidance

Identify the function: We have f(x) = $ \tan^{-1}(x^2) $. This is an inverse tangent function, which requires the use of the chain rule for differentiation.

Recall the derivative of $ \tan^{-1}(u) $: The derivative of $ \tan^{-1}(u) $ with respect to u is $ \frac{1}{1+u^2} $.

Apply the chain rule: Since we have $ u = x^2 $, we need to differentiate $ u $ with respect to x, which gives $ \frac{du}{dx} = 2x $.

Combine the derivatives: Using the chain rule, the derivative of $ f(x) = \tan^{-1}(x^2) $ is $ \frac{d}{dx}[\tan^{-1}(x^2)] = \frac{1}{1+(x^2)^2} \cdot 2x $.

Simplify the expression: The derivative becomes $ \frac{2x}{1+x^4} $ after simplifying $ 1+(x^2)^2 $ to $ 1+x^4 $.

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