Given the equation in picture below, what is? : a. #f'(x)# b. the derivative of #e^(f(x))#

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1 Answer
Steve M
May 10, 2018

(a) # f'(x) = -(5x+10)/(x(x-5)) #

(b) #d/dxe^f(x) = - (5ex( 2+x)) / (x-5)^8 #

Explanation:

We seek (a) #f'(x)#, and (b) #d/dxe^f(x)#, where

# f(x) = ln ((ex^2)/(x-5)^7) #

Part (a):

Using the properties of logarithms, we can write:

# f(x) = ln (e) + ln (x^2) - ln(x-5)^7 #

# \ \ \ \ \ \ \ = 1 +2lnx-7ln(x-5) #

So, we can differentiate to get:

# f'(x) = 0 +2/x -7/(x-5) #

# \ \ \ \ \ \ \ \ = (2(x-5)-7x)/(x(x-5)) #

# \ \ \ \ \ \ \ \ = (2x-10-7x)/(x(x-5)) #

# \ \ \ \ \ \ \ \ = -(5x+10)/(x(x-5)) #

Part (b):

Let:

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

Then:

# g(x) = e^(ln ((ex^2)/(x-5)^7)) #

# \ \ \ \ \ \ \ = (ex^2)/(x-5)^7 #

And using the quotient rule (and chain rule), we have:

# g'(x) = { ((x-5)^7)(2ex) - (7(x-5)^6)(ex^2) } / ((x-5)^7)^2 #

# \ \ \ \ \ \ \ \ = { 2ex(x-5) - 7ex^2 } / (x-5)^8 #

# \ \ \ \ \ \ \ \ = { 2ex^2-10ex - 7ex^2 } / (x-5)^8 #

# \ \ \ \ \ \ \ \ = { -10ex - 5ex^2 } / (x-5)^8 #

# \ \ \ \ \ \ \ \ = - (5ex( 2+x)) / (x-5)^8 #

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