Chapter 4.3, Problem 26E

To determine

To calculate:The equation of the tangent for the given function at the given point.

Answer to Problem 26E

The equation of tangent is $4y-4x-\pi +4=0$ .

Explanation of Solution

Given information:The function is $y={\tan }^{-1}(x^2)$ and the value of $x=1$ .

Formula used:

Slope-point form of line is:

  $\left(y-y_1\right)=m\left(x-x_1\right),\text{where}\left(x_1,y_1\right)\text{is}\text{the}\text{intercept}\text{and}m\text{is}\text{the}\text{slope}\text{of}\text{the}\text{line}$ .

  $\frac{d{\tan }^{-1}x}{dx}=\frac{1}{1+x^2}$

Also product, quotient and chain rule of derivatives are used.

Calculation:

Value of $x=1$ , in order to find the equation of the tangent the value of y and m is required to get the slope point form of the tangent.

The function is $y={\tan }^{-1}(x^2)$ ..........(1)

Put $x=1$ in equation (1)

  $\begin{array}{l}y={\tan }^{-1}(1^2)\\ y=\frac{\pi }{4}\text{radians}\end{array}$

By differentiating equation (1), we get

  $y={\tan }^{-1}(x^2)$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d{\tan }^{-1}\left(x^2\right)}{dx}\\ \frac{dy}{dx}=m=\frac{1}{1+{\left(x^2\right)}^2}\times 2x\mathrm{.........}\left(2\right)\end{array}$

Now put the value $x=1$ in equation (2)

  $\begin{array}{l}m=\frac{1}{1+{\left(1^2\right)}^2}\times 2\left(1\right)\\ m=\frac{1}{2}\times 2=1\end{array}$

Here $m=1$ is the slope of the tangent.

The slope point form of the tangent line is:

  $\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}$$m=1$ and $\left(x_1,y_1\right)=\left(1,\frac{\pi }{4}\right)$ .

  $\begin{array}{l}\left(y-\frac{\pi }{4}\right)=1\left(x-1\right)\\ \left(\frac{4y-\pi }{4}\right)=\left(x-1\right)\\ 4y-\pi =4x-4\\ 4y-4x-\pi +4=0\end{array}$

Hence the equation of tangent is $4y-4x-\pi +4=0$ .