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

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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)#