How do you differentiate # f(x)= [(2x-5)^5]/[(x^2 +2)^2] # using the chain rule.?

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Answer 1 · 2017-07-30T08:44:34

The derivative is #=(2(2x-5)^4(x^2+10x+10))/(x^2+3)^3#

This is the derivative of a quotient

#(u/v)'=(u'v-uv')/(v^2)#

Here,

#u(x)=(2x-5)^5#, #=>#, #u'(x)=5(2x-5)^4*2=10(2x-5)^4#

#v(x)=(x^2+2)^2#, #=>#, #v'(x)=2(x^2+2)*2x=4x(x^2+2)#

Therefore,

#f'(x)=(10(2x-5)^4(x^2+2)^2-(2x-5)^5(4x(x^2+2)))/(x^2+2)^4#

#=((x^2+2)(2x-5)^4(10(x^2+2)-4x(2x-5)))/(x^2+4)^4#

#=((2x-5)^4(10x^2+20-8x^2+20x))/(x^2+2)^3#

#=((2x-5)^4(2x^2+20x+20))/(x^2+2)^3#

#=(2(2x-5)^4(x^2+10x+10))/(x^2+3)^3#

Answer 2 · 2017-07-30T09:05:58

#(df)/dx = ( 2(2x-5)^4(x^2+10x+10))/ (x^2+2)^3#

First we need to use the quotient rule:

#(df)/dx = ( (x^2+2)^2 d/dx ((2x-5)^5) - (2x-5)^5 d/dx ( (x^2+2)^2))/ (x^2+2)^4#

Now using the chain rule:

#d/dx ((2x-5)^5) = 5(2x-5)^4 d/dx (2x-5) = 10(2x-5)^4#

#d/dx ( (x^2+2)^2)) = 2(x^2+2)d/dx (x^2+2) = 4x(x^2+2)#

so:

#(df)/dx = ( 10(x^2+2)^2 (2x-5)^4 - 4x (2x-5)^5 (x^2+2))/ (x^2+2)^4#

#(df)/dx = ( 10(x^2+2) (2x-5)^4 - 4x (2x-5)^5 )/ (x^2+2)^3#

#(df)/dx = ( (2x-5)^4(10(x^2+2) - 4x (2x-5) ))/ (x^2+2)^3#

#(df)/dx = ( (2x-5)^4(10x^2+20 - 8x^2+20x ))/ (x^2+2)^3#

#(df)/dx = ( (2x-5)^4(2x^2+20x+20))/ (x^2+2)^3#