How do you find the derivative of #f(x)=2/(6x+5x+1)^2#?

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Answer 1 · 2018-03-13T21:15:09

I am a bit old fashioned in that I prefer #dy/dx #
Should the #6x# be #6x^2 larr" More logical"#

#f'(x)=dy/dx=-(4(12x+5))/(6x^2+5x+1)^3#

Given: #f(x)=2/(6x^2+5x+1)^2#

This is the same as #f(x)=2(6x^2+5x+1)^(-2)#

Set #u=6x^2+5x+1 => (du)/dx=12x+5#

Set #y=2u^(-2) => dy/(du)=-4u^(-3)#

#dy/dx=(du)/dx xx dy/(du)#

#dy/dx =(12x+5)xx (-4u^(-3))#

#dy/dx=-(4(12x+5))/(6x^2+5x+1)^3#

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Solution Ex Maple matches

#-(4(12x+5))/(6x^2+5x+1)^3#