How do you find the derivative of #x^tanx#?

1 Answer
Bill K.
Aug 4, 2015

#x^{tan(x)}(ln(x)*sec^{2}(x)+tan(x)/x)#

Explanation:

Use logarithmic differentiation: let #y=x^{tan(x)}# so that #ln(y)=ln(x^{tan(x)})=tan(x)ln(x)#.

Now differentiate both sides with respect to #x#, keeping in mind that #y# is a function of #x# and using the Chain Rule and Product Rule:

#1/y * dy/dx=sec^{2}(x)ln(x)+tan(x)/x#

Hence,

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

#=x^{tan(x)} (ln(x)sec^{2}(x)+tan(x)/x)#

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