Find the derivative of $y=(\tan (x))^{\log (x)}$

Question

Find the derivative of $y=(\tan (x))^{\log (x)}$

I thought of using the power rule that: $$\dfrac {d}{dx} u^n = n.u^{n-1}.\dfrac {du}{dx}$$ Realizing that the exponent $log(x)$ is not constant, I could not use that.

Answer 1

$$\left(\left(\tan{x}\right)^{\ln{x}}\right)'=\left(e^{\ln{x}\ln\tan{x}}\right)'=e^{\ln{x}\ln\tan{x}}\left(\ln{x}\ln\tan{x}\right)'=$$ $$=\left(\tan{x}\right)^{\ln{x}}\left(\frac{\ln\tan{x}}{x}+\frac{\ln{x}}{\tan{x}}\cdot\frac{1}{\cos^2x}\right)=\left(\tan{x}\right)^{\ln{x}}\left(\frac{\ln\tan{x}}{x}+\frac{2\ln{x}}{\sin2x}\right).$$

Answer 2

Note that $$\log(y) = \log(x).\log(\tan(x))~.$$ Now, apply chain rule on both sides.

Answer 3

$(\tan(x))^{\ln(x)}=\exp(\ln x(\ln(\tan x)))$ by apply the formula off the derivative of a composition:

$f'(x)=(\ln(x)\ln(\tan(x))'\exp(\ln(\tan(x))$.

$(\ln(x)(\ln(\tan(x)))'={1\over x}\ln(\tan(x))+\ln(x){1\over {\tan(x)}}{1\over{cos(x)^2}}$.

Answer 4

You can rewrite $y = (\tan{(x)})^{\log{(x)}}$ as $$y=\exp\left(\log{(\tan{(x)})}\log(x)\right)$$ using the rule $a^b = \exp(\log(a))^b = \exp{(b\cdot\log{(a)})}$.

In this form, you can find the derivative of $y$ using the chain rule and product rule.