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

Question

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

Cameron Alexander · Accepted Answer

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

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)#

Christopher Allers · Answer

#-5e^(5x)# #"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)#

Scarlett Allison · Answer

undefined To differentiate $ f(g(x)) $ using the chain rule, first find the derivative of the outer function $ f(x) $ with respect to its inner function $ g(x) $, then multiply it by the derivative of the inner function $ g(x) $ with respect to $ x $.

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

1. Find $ f'(x) $, the derivative of $ f(x) $ with respect to $ x $: $ f'(x) = -e^x $.
2. Find $ g'(x) $, the derivative of $ g(x) $ with respect to $ x $: $ g'(x) = 5 $.
3. Substitute $ g(x) $ into $ f'(x) $ to get $ f'(g(x)) $: $ f'(g(x)) = -e^{g(x)} $.
4. Multiply $ f'(g(x)) $ by $ g'(x) $: $ f'(g(x)) \cdot g'(x) = -5e^{g(x)} $.

So, $ \frac{d}{dx}[f(g(x))] = -5e^{g(x)} $.