Question

Logarithmic differentiation of (tan(3x^2+3sin(x^4)))^(3y^2) help anyone?

Answer 1

Let L=(tan(3x^2+3sin(x^4)))^(3y^2)
LogL=(3y^2)*Log(tan(3x^2+3sin(x^4)))
diferrentiate on both sides,
1/L*dL=(6ydy)Log(tan(3x^2+3sin(x^4))) +3y^2*[1/(tan(3x^2+3sin(x^4))]*[sec(3x^2+3sin(x^4))]^2*[6xdx+[3cos(x^4)*(4x^3)]dx]
Now what we had to find initially was dL
dL=L*{(6ydy)Log(tan(3x^2+3sin(x^4))) +3y^2*[1/(tan(3x^2+3sin(x^4))]*[sec(3x^2+3sin(x^4))]^2*[6xdx+[3cos(x^4)*(4x^3)]dx]}
therefore
dL=(tan(3x^2+3sin(x^4)))^(3y^2)*{(6ydy)Log(tan(3x^2+3sin(x^4))) +3y^2*[1/(tan(3x^2+3sin(x^4))]*[sec(3x^2+3sin(x^4))]^2*[6xdx+[3cos(x^4)*(4x^3)]dx]}
I hope i didnt make any mistakes because it is quite a long differentition to do on the keyboard!