Chapter 3.4, Problem 41E

To determine

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

Answer to Problem 41E

The equation of the tangent line is $y=20x+1$ .

Explanation of Solution

Given:

The given curve is $y={\left(1+2x\right)}^{10}$ and the given point is $\left(0,1\right)$ .

Calculation:

Find the slope of the tangent line at the given point $\left(0,1\right)$ .

Find first derivative of the given equation $y={\left(1+2x\right)}^{10}$ .

Apply chain rule.

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

Let $f=a^{10},a=1+2x$

  $\Rightarrow y\text{'}=\frac{d}{da}\left(a^{10}\right)\cdot \frac{d}{dx}\left(1+2x\right)$

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

  $\begin{array}{l}\Rightarrow y\text{'}=10a^9\cdot \left(0+2\right)\\ \Rightarrow y\text{'}=20a^9\end{array}$

Substitute the value of $a=1+2x$ .

  $\Rightarrow y\text{'}=20{\left(1+2x\right)}^9$

Plug in the $x=0$ and $y=1$ into the derivative.

  $\Rightarrow y\text{'}=20{\left(1+2\cdot 0\right)}^9=20$

Slope of the tangent line is $20$ .

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

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y_1=1,x_1=0,m=20\\ y-1=20\left(x-0\right)\\ \Rightarrow y-1=20x\\ \Rightarrow y-1+1=20x+1\\ \Rightarrow y=20x+1\end{array}$

Hence theequation of the tangent line is $y=20x+1$ .