Question

In Exercises \(23-26\) , find an equation for the tangent to the graph of \(y\) at the indicated point. $$y=\tan ^{-1}\left(x^{2}\right), \quad x=1$$

Short answer

The equation of the tangent to the graph of \(y = \tan^{-1}(x^2)\) at the point where \(x = 1\) is \(y = x + \frac{\pi}{4} - 1\).

Step-by-step solution

Find the Derivative of Given Function

First, find the derivative of the given function using the chain rule. The derivative of \(y = \tan^{-1}(x^2)\) is \(y' = \frac{2x}{1+(x^2)^2}\).

Calculate the Slope at Given Point

Substitute \(x = 1\) into the derivative to get the slope of the tangent line at the point. The slope is \(m = \frac{2(1)}{1+(1)^2} = \frac{2}{2} = 1\).

Find the Value of Function at the Given Point

Substitute \(x = 1\) into the original function to get the y-coordinate of the point. So, \(y = \tan^{-1}(1^2) = \tan^{-1}(1) = \frac{\pi}{4}\). Now we have the point \((1, \frac{\pi}{4})\).

Write the Equation of the Tangent Line

The form of the equation of a line is \(y = mx + b\). We have the slope \(m = 1\) and the point \((1, \frac{\pi}{4})\), we substitute these values into the line equation to solve for b: \(\frac{\pi}{4} = 1 \times 1 + b\), so \(b = \frac{\pi}{4} - 1\). Then, our equation of the tangent line is \(y = x + \frac{\pi}{4} - 1\).