Question #93d8f

2 Answers
Jan 6, 2018

#dy/dx=e^(x+1)#

Explanation:

.

#y=e^(x+1)#

Let #u=x+1#

#(du)/dx=1#

#y=e^u#

#dy/(du)=e^u#

Using the Chain Rule:

#dy/dx=dy/(du)*(du)/dx#

#dy/dx=e^u(1)#

#dy/dx=e^(x+1)#

Jan 6, 2018

#e^(x+1)#

Explanation:

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

#"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)#