How to differentiate y= (ln(4x^2))/(3^(6x)y=ln(4x2)36x ?

1 Answer
Ananda Dasgupta
May 29, 2018

(2-6x ln 3 ln(4x^2))/(3^(6x)x)2−6xln3ln(4x2)36xx

Explanation:

You can use the quotient rule

d/dx (u/v) = (v (du)/dx - u(dv)/dx)/v^2ddx(uv)=vdudx−udvdxv2

Here

u = ln(4x^2) = ln 4+2ln x impliesu=ln(4x2)=ln4+2lnx⇒
(du)/dx = 2/xdudx=2x

and

v = 3^(6x) = (e^{ln 3})^{6x} = e^{6 (ln 3 )x}impliesv=36x=(eln3)6x=e6(ln3)x⇒

(dv)/dx = 6 ln 3 times e^{6 (ln 3 )x} = 6 ln 3 times 3^(6x)dvdx=6ln3×e6(ln3)x=6ln3×36x

Hence

d/dx(ln(4x^2)/3^(6x)) =(3^(6x)times 2/x-ln(4x^2)times 6 ln 3 times 3^(6x))/(3^(6x))^2 ddx(ln(4x2)36x)=36x×2x−ln(4x2)×6ln3×36x(36x)2
qquad = (2-6x ln 3 ln(4x^2))/(3^(6x)x)

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