How do you differentiate $y=(lnx)^tanx$ ?

Question

12597 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-09T17:30:51

Use logarithmic differentiation.

Use the natural logarithm on both sides:

$ln(y) = ln(ln(x)^tan(x))$

Use the property of logarithms $ln(a^b) = (b)ln(a)$

$ln(y) = tan(x)ln(ln(x))$

Differentiate:

$1/y(dy/dx) = sec^2(x)ln(ln(x)) + tan(x)/(xln(x))$

Multyply both sides by y:

$dy/dx = {sec^2(x)ln(ln(x)) + tan(x)/(xln(x))}y$

Substitute $ln(x)^tan(x)$ for y:

$dy/dx = {sec^2(x)ln(ln(x)) + tan(x)/(xln(x))}ln(x)^tan(x)$

How do you differentiate y=(lnx)^tanxy=(lnx)^tanx?