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

1 Answer
Jan 14, 2017

#"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#