How to find d/dx?

Question

d/dx {(1+x^2+x^4)^2x-3}

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Answer 1 · 2018-02-17T08:49:50

#(dy)/(dx)=[2ln(1+x^2+x^4)+(8x^4-12x^3+4x^2-6x)/(1+x^2+x^4)](1+x^2+x^4)^(2x-3)#

Let #y=(1+x^2+x^4)^(2x-3)#

i.e. #lny=(2x-3)ln(1+x^2+x^4)#

and then #1/y(dy)/(dx)=2ln(1+x^2+x^4)+(2x-3)/(1+x^2+x^4)(2x+4x^3)#

= #2ln(1+x^2+x^4)+(8x^4-12x^3+4x^2-6x)/(1+x^2+x^4)#

and #(dy)/(dx)=[2ln(1+x^2+x^4)+(8x^4-12x^3+4x^2-6x)/(1+x^2+x^4)](1+x^2+x^4)^(2x-3)#