If #f(x)= - e^x # and #g(x) = 5 x #, how do you differentiate #f(g(x)) # using the chain rule?

2 Answers
Dec 4, 2017

Answer:

#-5e^(5x)#

Explanation:

#"to evaluate "f(g(x)" substitute "g(x)" into "f(x)#

#rArrf(g(x)=f(5x)=-e^(5x)#

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

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#d/dx[f(g(x))]#

#=-e^(5x)xxd/dx(5x)=-5e^(5x)#

Dec 4, 2017

Answer:

#d/dx(f(g(x)))=-5e^(5x)#

Explanation:

Given #f(x)=-e^x# and #g(x)=5x#

So,

#f(g(x))=-e^(g(x))#

As #g(x)=5x#

Therefore,

#f(g(x))=-e^(5x)#

Now differentiate both sides with respect to #x# using the chain rule.

The chain rule says that #->#

#(U(V(x)))'=U'(V(x)) xx V'(x)#

Now back to the question #->#

#f(g(x))=-e^(5x)#

#d/dx(f(g(x)))=d/dx-e^(5x)#

#d/dx(f(g(x)))=-e^(5x) xx d/dx 5x#

#d/dx(f(g(x)))=-e^(5x) xx 5#

Therefore,

#d/dx(f(g(x)))=-5e^(5x)#