Question

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

Answer 1

`"slope " =dy/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 `color(blue)"chain rule"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(2/2)|)))`

`"let " u=5x^2+7rArr(du)/(dx)=10x`

`"then " y=u^2rArr(dy)/(du)=2u`

`rArrdy/dx=2u.10x=20xu`

change u back into terms of x.

`rArrdy/dx=20x(5x^2+7)`

`x=1tody/dx=20(5+7)=240`