Question

What is the derivative of ${{\left( \cos x \right)}^{\sin x}}$?

Answer 1

Assume the given function as $y=f\left( x \right)$. Take log to the base e both the sides and use the property of logarithm given as $\ln {{a}^{m}}=m\ln a$ to simplify the R.H.S. Now, differentiate both the sides of the assumed function y and use the chain rule of derivative to find the derivative of L.H.S. Use the formulas $\dfrac{d\left( \ln y \right)}{dy}=\dfrac{1}{y}$, $\dfrac{d\left( \sin x \right)}{dx}=\cos x$ and $\dfrac{d\left( \cos x \right)}{dx}=-\sin x$ for the simplification. For the derivative of R.H.S use the product rule of differentiation given as $\dfrac{d\left( u\times v \right)}{dx}=v\dfrac{du}{dx}\times u\dfrac{dv}{dx}$ where u and v are functions of x. Finally, substitute back the assumed value of y to get the value of $\dfrac{dy}{dx}$.

Complete step by step answer:
Here we have been provided with the function ${{\left( \cos x \right)}^{\sin x}}$ and we are asked to differentiate it. Let us assume this function as y so we have,
$\Rightarrow y={{\left( \cos x \right)}^{\sin x}}$
Now, we need to find the value of $\dfrac{dy}{dx}$. Taking natural log, i.e. log to the base e, on both the sides we get,
$\Rightarrow \ln y=\ln \left( {{\left( \cos x \right)}^{\sin x}} \right)$
Using the property of log given as $\ln {{a}^{m}}=m\ln a$ we get,
$\Rightarrow \ln y=\sin x\ln \left( \cos x \right)$
Differentiating both the sides with respect to x we get,
$\Rightarrow \dfrac{d\ln y}{dx}=\dfrac{d\left( \sin x\times \ln \left( \cos x \right) \right)}{dx}$
Using the chain rule of derivative in the L.H.S where we will differentiate $\ln y$ with respect to y and then its product will be taken with the derivative of y with respect to x, so we get,
$\Rightarrow \dfrac{d\ln y}{dy}\times \dfrac{dy}{dx}=\dfrac{d\left( \sin x\times \ln \left( \cos x \right) \right)}{dx}$
Using the formula $\dfrac{d\left( \ln y \right)}{dy}=\dfrac{1}{y}$ we get,
$\Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}=\dfrac{d\left( \sin x\times \ln \left( \cos x \right) \right)}{dx}$
Using the product rule of derivative given as $\dfrac{d\left( u\times v \right)}{dx}=v\dfrac{du}{dx}\times u\dfrac{dv}{dx}$ assuming $u=\sin x$ and $v=\ln \left( \cos x \right)$ in the R.H.S we get,
$\Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}=\sin x\times \dfrac{d\left( \ln \left( \cos x \right) \right)}{dx}+\ln \left( \cos x \right)\times \dfrac{d\left( \sin x \right)}{dx}$
Using the formulas $\dfrac{d\left( \sin x \right)}{dx}=\cos x$ and chain rule of derivative for the differentiation of $\ln \left( \cos x \right)$ we get,
$\Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}=\sin x\times \dfrac{d\left( \ln \left( \cos x \right) \right)}{d\left( \cos x \right)}\times \dfrac{d\left( \cos x \right)}{dx}+\ln \left( \cos x \right)\times \cos x$
Using the formula $\dfrac{d\left( \cos x \right)}{dx}=-\sin x$ we get,
$\begin{aligned} & \Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}=\sin x\times \dfrac{1}{\left( \cos x \right)}\times \left( -\sin x \right)+\ln \left( \cos x \right)\times \cos x \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{y}\left[ \cos x\ln \left( \cos x \right)-\dfrac{{{\sin }^{2}}x}{\cos x} \right] \\\ \end{aligned}$
Substituting back the value of y we get,
$\therefore \dfrac{dy}{dx}=\dfrac{1}{{{\left( \cos x \right)}^{\sin x}}}\left[ \cos x\ln \left( \cos x \right)-\dfrac{{{\sin }^{2}}x}{\cos x} \right]$
Hence, the above relation is our answer.

Note: You can also remember the direct formula for the derivative of the function of the form ${{u}^{v}}$ where u and v both are variables and functions of x. What we have to do is we will consider two parts of the derivative. In the first part we will assume u as constant and v as derivative and differentiate using the formula $\dfrac{d\left( {{u}^{v}} \right)}{dx}={{u}^{v}}\log u\left( \dfrac{dv}{dx} \right)$. In the second part we will consider v as constant and u as variable and apply the formula $\dfrac{d\left( {{u}^{v}} \right)}{dx}=v{{u}^{v-1}}\left( \dfrac{du}{dx} \right)$ for the derivative. Finally, we will take the sum of these two expressions to get the answer.