How would I differentiate $\sin{x}^{\cos{x}}?$

How can I differentiate $\displaystyle \sin{x}^{\cos{x}}$? I know the power rule will not work in this case, but logarithmic differentiation would. I’m not sure how to start the problem though and I’m not too comfortable with logarithmic differentiation.

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Let $y = \sin x ^{\cos x } $ , then

$$ \ln y = \cos x \ln (\sin x ) \implies \frac{y’}{y} = -\sin x \ln(\sin x) + \cos x\frac{\cos x}{\sin x}$$

$$ \therefore y’ = \sin x ^{\cos x } \left( \frac{\cos^2 x}{\sin x} – \sin x \ln(\sin x)\right) $$

In general,

Let $f, g$ be any functions. Let $y = f^g \implies \ln y = g \ln f $

$$ \therefore \frac{y’}{y} = g’ \ln f + g\frac{f’}{f} \implies \frac{ df^g}{dx}= y’ = f^g \left( f’ \ln f + \frac{g f’}{f} \right)$$

In addition to the other answers, you may also rewrite the function as

$$[e^{\ln(\sin x)}]^{\cos x}=e^{\ln(\sin x)\cos x}$$

which you should be able to take the derivative of through the chain rule and product rule.

Let $f(x) = sinx^{cosx}$ then $log(f(x)) = cosx log sinx $. Now differebtiate LHS by function of function rule and RHS bu product rule and finally solve for $f(x)$.

Consider $$y=\sin(x)^{\cos(x)}.$$ We will use the method of logarithmic differentiation to obtain this functions derivative. Take the natural logarithm of both sides of the equation and use the properties of logarithms to simplify. So $$\ln(y)=\cos(x)\cdot \ln(\sin(x)).$$ Differentiating implicitly with respect to $x$ we obtain $${1\over y}\cdot y’=\cos(x)\cdot {\cos(x)\over \sin(x)}-\sin(x)\cdot \ln(\sin(x)).$$ Simplifying the right-hand side we see that $${1\over y}\cdot y’=\cos(x)\cdot \cot(x)-\sin(x)\cdot \ln(sin(x)).$$ Multiply both sides of the equation by $y$ and we have $$y’=\sin(x)^{\cos(x)}(\cos(x)\cdot \cot(x)-\sin(x)\cdot \ln(\sin(x)).$$