We read your photo as:

(3 points) Show that d/dx [tan⁻¹ x] = 1/(1+x2)

June 7, 2026Mathematics

ScanSolve AI Answer

Answer

\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}

5 steps

To show that $\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}$ , we can use implicit differentiation.

Let $y = \tan^{-1}(x)$ .
This means that $x = \tan(y)$ .
Differentiate both sides of $x = \tan(y)$ with respect to $x$ .
We use the chain rule for the right side.
$\frac{d}{dx}(x) = \frac{d}{dx}(\tan(y))$ $1 = \sec^2(y) \frac{dy}{dx}$
Solve for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{1}{\sec^2(y)}$
Use the trigonometric identity $\sec^2(y) = 1 + \tan^2(y)$ .
Substitute this into the expression for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{1}{1 + \tan^2(y)}$
Substitute $x$ back into the equation, since we know $x = \tan(y)$ .
$\frac{dy}{dx} = \frac{1}{1 + x^2}$

Was this helpful?

A new problem that tests the same skill.

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.