Question #93d8f

2 Answers
Jan 6, 2018

dy/dx=e^(x+1)dydx=ex+1

Explanation:

.

y=e^(x+1)y=ex+1

Let u=x+1u=x+1

(du)/dx=1dudx=1

y=e^uy=eu

dy/(du)=e^udydu=eu

Using the Chain Rule:

dy/dx=dy/(du)*(du)/dxdydx=dydududx

dy/dx=e^u(1)dydx=eu(1)

dy/dx=e^(x+1)dydx=ex+1

Jan 6, 2018

e^(x+1)ex+1

Explanation:

"differentiate using the "color(blue)"chain rule"differentiate using the chain rule

"given "y=f(g(x))" then "given y=f(g(x)) then

dy/dx=f'(g(x))xxg'(x)

rArrd/dx(e^(x+1))=e^(x+1)xxd/dx(x+1)=e^(x+1)