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

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Answer 1 · 2016-02-07T09:48:56

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

The chain rule states that

#d/dx f(g(x)) = f'(g(x))g'(x)#

In this case

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

To find the derivative of #g(x)# we must again use the chain rule on

#g(x) = g_1((g_2(x))# where #g_1(x) = sqrt(x)# and #g_2(x)=1-x#

#g_1'(x) = 1/(2sqrt(x))#

#g_2'(x) = -1#

#=> g'(x) = d/dxg_1(g_2(x)) = g_1'(g_2(x))g_2'(x) = -1/(2sqrt(1-x))#

Thus, putting it together

#d/dx f(g(x)) = f'(g(x))g'(x)#

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

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