If #f(x)= - e^x # and #g(x) =1 / sqrt(1-x #, how do you differentiate #f(g(x)) # using the chain rule?

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Answer 1 · 2017-11-22T01:04:32

Substitute g in for x to find f(g) from f(x), and recall #d/dx f(g(x))= (df)/(dg) (dg)/dx#. See explanation.

Recall that the chain rule states that if #y=f(g(x)), dy/dx =(df)/(dg)*(dg)/(dx)#

We have #f(x) =-e^x, g(x) = 1/sqrt(1-x)#

(This step is unnecessary, but in case the student is curious, to find f(g) you substitute g in for x in f(x), making #f(g) = -e^g#

We know that #d/(du) ke^u=ke^u#, and #d/dx(1/(1-x)^(1/2)) = 1/2 * 1/(1-x)^(3/2) = 1/(2(1-x)^(3/2))#.
Thus...

#dy/dx = (df)/(dg) (dg)/(dx)= 1/(2(1-x)^(3/2)) *- e^(1/sqrt(1-x)) = (-e^(1/sqrt(1-x)))/(2(1-x)^(3/2))#

Answer 2 · 2017-11-22T01:17:12

#d/dx f(g(x)) = - (e^(sqrt(1-x)))/(2(1-x)^(3/2))#

#f(x) = -e^x and g(x) = 1/sqrt(1-x)#

Replace #x# in #f(x)# with g(x)#

#-> f(g(x)) = -e^(1/sqrt(i-x))#

#= -e^((1-x)^(-1/2))#

Apply the chain rule and standard differential

#d/dx f(g(x)) = -e^((1-x)^(-1/2)) * d/dx (1-x)^(-1/2)#

Apply the chain rule and power rule

#d/dx f(g(x)) = -e^((1-x)^(-1/2)) * (-1/2)(1-x)^(-3/2)* d/dx (1-x)#

#= -e^((1-x)^(-1/2)) * (-1/2)(1-x)^(-3/2)* (-1)#

#= -e^(1/sqrt(1-x)) * 1/(2(1-x)^(3/2))#

#= - (e^(sqrt(1-x)))/(2(1-x)^(3/2))#