Chapter 3.4, Problem 42E

To determine

Find the equation of the tangent line for the given curve.

Answer to Problem 42E

The equation of the tangent line is $y=2x-1$ .

Explanation of Solution

Given:

The given curve is $y=\sqrt{1+x^3}$ and the given point is $\left(2,3\right)$ .

Calculation:

Find the slope of the tangent line at the given point $\left(2,3\right)$ .

Find first derivative of the given equation $y=\sqrt{1+x^3}$ .

Apply chain rule.

  $\frac{df\left(a\right)}{dx}=\frac{df}{da}\cdot \frac{da}{dx}$

Let $f=a^{\frac{1}{2}},a=1+x^3$

  $\Rightarrow y\text{'}=\frac{d}{da}\left(a^{\frac{1}{2}}\right)\cdot \frac{d}{dx}\left(1+x^3\right)$

Use derivative rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\Rightarrow y\text{'}=\frac{3x^2}{2\sqrt{a}}$

Substitute the value of $a=1+x^3$ .

  $\Rightarrow y\text{'}=\frac{3x^2}{2\sqrt{1+x^3}}$

Plug in the $x=2$ and $y=3$ into the derivative.

  $\Rightarrow y\text{'}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}=\frac{12}{6}=2$

Slope of the tangent line is $2$ .

Use point-slope form of the equation (For the equation of tangent line).

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y_1=3,x_1=2,m=2\\ y-3=2\left(x-2\right)\\ \Rightarrow y-3=2x-4\\ \Rightarrow y-3+3=2x-4+3\\ \Rightarrow y=2x-1\end{array}$

Hence theequation of the tangent line is $y=2x-1$ .