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

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Answer 1 · 2015-08-04T20:12:50

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

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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