# How do you find the slope of the tangent line y=(5x^2+7)^2 at x=1?

Jan 14, 2017

$\text{slope } = \frac{\mathrm{dy}}{\mathrm{dx}} = 240$

#### Explanation:

The slope of the tangent to the function y at x = 1 is the derivative of the function evaluated at x = 1

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{let } u = 5 {x}^{2} + 7 \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 10 x$

$\text{then } y = {u}^{2} \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = 2 u$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 2 u .10 x = 20 x u$

change u back into terms of x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 20 x \left(5 {x}^{2} + 7\right)$

$x = 1 \to \frac{\mathrm{dy}}{\mathrm{dx}} = 20 \left(5 + 7\right) = 240$