Chapter 3.7, Problem 27E

To determine

To calculate: The equation of the tangent line to the curve at the given point.

Answer to Problem 27E

The equation of the tangent line to the curve is $y=3x-2$ .

Explanation of Solution

Given information:

The function $y=\ln \left(x{e}^{x^2}\right)$ and the point $\left(1,1\right)$ .

Formula used:

Chain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Logarithmic rule for differentiation is $\frac{d}{dx}\log x=\frac{1}{x}$ .

Equation of tangent to any curve at any point $\left(x_1,y_1\right)$ is given by $\left(y-y_1\right)={\frac{dy}{dx}]}_{\left(x_1,y_1\right)}\left(x-x_1\right)$ .

Calculation:

Consider the function $y=\ln \left(x{e}^{x^2}\right)$ and the point $\left(1,1\right)$ .

Differentiate curve both sides with respect to xto determine the slope of tangent line,

  $\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\ln \left(x{e}^{x^2}\right)\right)$

Recall that power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ and logarithmic rule for differentiation is $\frac{d}{dx}\log x=\frac{1}{x}$ .

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\ln \left(x{e}^{x^2}\right)\right)\\ \frac{dy}{dx}=\frac{1}{x{e}^{x^2}}\cdot \left(e^{x^2}+x{e}^{x^2}\left(2x\right)\right)\\ \frac{dy}{dx}=\frac{e^{x^2}\left(1+2x^2\right)}{x{e}^{x^2}}\\ \frac{dy}{dx}=\frac{1+2x^2}{x}\end{array}$

Slope of tangent at the point $\left(1,1\right)$ is calculated as,

  $\begin{array}{c}\frac{dy}{dx}=\frac{1+2x^2}{x}\\ {\frac{dy}{dx}]}_{\left(1,1\right)}=\frac{1+2\left(1^2\right)}{1}\\ =\frac{1+2}{1}\\ =3\end{array}$

Recall that the equation of tangent to any curve at any point $\left(x_1,y_1\right)$ is given by $\left(y-y_1\right)={\frac{dy}{dx}]}_{\left(x_1,y_1\right)}\left(x-x_1\right)$ .

  $\begin{array}{c}\left(y-y_1\right)={\frac{dy}{dx}]}_{\left(x_1,y_1\right)}\left(x-x_1\right)\\ y-1=3\left(x-1\right)\\ y-1=3x-3\\ y=3x-2\end{array}$

Thus, the equation of the tangent line to the curve is $y=3x-2$ .