Question

Use logarithmic differentiation to find y'?

`((x^3 +1)^4 . e^(5x^2) . (1+ tan^2 x))/(sqrt(1+2x^4) . cot^2 .(x^2+1))`##

Answer 1

`dy/dx = ({12x^2}/(x^3+1)+10x+2 tan(x)- {4x^3}/(1+2x^4)+4xcsc(x^2+1)sec(x^2+1))y`

Explanation:

`y = ((x^3 +1)^4 . e^(5x^2) . (1+ tan^2 x))/(sqrt(1+2x^4) . cot^2 .(x^2+1)) implies`

`ln y = 4 ln(x^3+1)+5x^2+2 ln |sec(x)|-1/2 ln(1+2x^4)-2ln|cot(x^2+1)|`

Differentiating both sides with respect to `x`

`1/y dy/dx = 4{3x^2}/(x^3+1)+10x+2 {sec(x)tan(x)}/sec(x)-1/2 {8x^3}/(1+2x^4)-2{-csc^2(x^2+1)2x}/cot(x^2+1)`

Simplifying somewhat

`1/y dy/dx = {12x^2}/(x^3+1)+10x+2 tan(x)- {4x^3}/(1+2x^4)+4xcsc(x^2+1)sec(x^2+1)`