How do you find the derivative of (2e^(3x) + 2e^(-2x))^4?
1 Answer
Aug 14, 2017
Explanation:
"differentiate using the "color(blue)"chain rule"
"given "y=f(g(x))" then"
dy/dx=f'(g(x))xxg'(x)larr" chain rule"
f(g(x))=(2e^(3x)+2e^(-2x))^4
rArrf'(g(x))=4(2e^(3x)+2e^(-2x))^3
g(x)=2e^(3x)+2e^(-2x)
rArrg'(x)=2e^(3x).d/dx(3x)+2e^(-2x).d/dx(-2x)
color(white)(rArrg'(x))=6e^(3x)-4e^(-2x)
rArrdy/dx=4(2e^(3x)+2e^(-2x))^3(6e^(3x)-4e^(-2x))