what is the derivative of (sinx)^tanx??
logarithmic differentiation ...
$\displaystyle y = (\sin{x})^{\tan{x}}$
$\displaystyle \ln(y) = \ln\left[(\sin{x})^{\tan{x}}\right]$
$\displaystyle \ln(y) = \tan{x} \cdot \ln(\sin{x})$
take the derivative w/r to x ...
$\displaystyle \frac{y'}{y} = \tan{x} \cdot \frac{\cos{x}}{\sin{x}} + \ln(\sin{x}) \cdot \sec^2{x}$
$\displaystyle \frac{y'}{y} = 1 + \sec^2{x} \cdot \ln(\sin{x})$
$\displaystyle y' = (\sin{x})^{\tan{x}}\left[1 + \sec^2{x} \cdot \ln(\sin{x})\right]$
$\displaystyle sin(x)^{tan(x)} = y$
$\displaystyle log(y) = tan(x)log(sin(x)) $
$\displaystyle
\frac{1}{y} \frac{dy}{dx} = sec^2(x)log(sin(x)) + tan(x)\times cos(x)\times \frac{1}{sin(x)} $ ------Using Chain Rule
$\displaystyle
\frac{dy}{dx} = y\times (sec^2(x)log(sin(x)) + 1) $
If you want put the value of y in RHS that's your answer
EDIT: Skeeter won
why did you not mention that to begin with?
$\displaystyle y = x^{\sin{x}^{\tan{x}}}$
$\displaystyle \ln{y} = \ln\left(x^{\sin{x}^{\tan{x}}}\right)$
$\displaystyle \ln{y} = \sin{x}^{\tan{x}} \cdot \ln{x}$
use the product rule ...
$\displaystyle \frac{y'}{y} = \sin{x}^{\tan{x}} \cdot \frac{1}{x} + \ln{x} \cdot (derivative \, from \, my \, previous \, post)$
finish up.
Yes, you can, using the chain rule.
By the way, here's an interesting point: if $\displaystyle y= f(x)^{g(x)}$, where f and g are integrable functions, there are two common mistakes in finding the derivative:
1) Ignore the fact that g is a variable and use the power law:
$\displaystyle y'= g(x) f(x)^{g(x)-1}$
2) Ignore the fact that f is a variable and treat it as an exponential:
$\displaystyle y'= ln(f(x)) f(x)^{g(x)}$
As I said, those are both wrong. The correct derivative is the sum of those!
$\displaystyle y'= g(x) f(x)^{g(x)-1}+ ln(f(x)) f(x)^{g(x)}$