What is the derivative of cos(x)^ln(x)? When I plug it into mathematica, I don't know where the answer comes from...
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Originally Posted by soma What is the derivative of cos(x)^ln(x)? When I plug it into mathematica, I don't know where the answer comes from... I would approach this by taking the logarithm of both sides. $\displaystyle ln(y)=ln(cos^{ln(x)}x)$ $\displaystyle ln(y)=ln(x)lncos(x)$ The use implicit differentiation.
So I get e^((ln(cos(x))/x)-ln(x)tan(x)) But mathematica shows (cos(x)^ln(x))*((ln(cos(x))/x)-ln(x)tan(x)) The difference is I have e^, while mathematica has (cos(x)^ln(x))*
I get: $\displaystyle \frac{1}{y}\frac{dy}{dx} =\frac{1}{x}lncos(x) -ln(x)tan(x)$ $\displaystyle \frac{dy}{dx}=\frac{y}{x}lncos(x)-yln(x)tan(x)$
Oh! don't forget that y=$\displaystyle cos^{ln(x)}(x)$ $\displaystyle \frac{dy}{dx}=cos^{ln(x)}(x)[\frac{lncos(x)}{x}-ln(x)tan(x)]$
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