$x$ is an algebraic function and $\sin{x}$ is a trigonometric function. The two functions formed a special function in exponential form and it is denoted by a variable $y$.

$y \,=\, x^{\,\displaystyle \sin{x}}$

In this differentiation problem, we have to evaluate the derivative of the function $y$ with respect to $x$.

The function $y$ is in exponential form and it is not easy to differentiate such functions. Hence, we have to either remove or reduce the complexity of the function. It can be done by the logarithmic operation. Therefore, take natural logarithm both sides of the equation.

$\implies$ $\log_{e}{y}$ $\,=\,$ $\log_{e}{\Big(x^{\,\displaystyle \sin{x}}\Big)}$

The logarithmic expression at the right hand side of the equation can be simplified by the power rule of logarithms.

$\implies$ $\log_{e}{y}$ $\,=\,$ $\sin{x} \times \log_{e}{x}$

The function $y$ is defined in $x$. So, let’s differentiate the equation with respect to $x$ for evaluating the derivative of $y$ with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\,\log_{e}{y}}$ $\,=\,$ $\dfrac{d}{dx}{\,\Big(\sin{x} \times \log_{e}{x}\Big)}$

In this differentiation problem, the variable $y$ represents a function in $x$. Hence, it can be differentiated with respect to $x$ and do not think that $y$ is a constant. Therefore, the function $y$ can be differentiated by the derivative rule of logarithms.

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{\,\Big(\sin{x} \times \log_{e}{x}\Big)}$

The functions $\sin{x}$ and $\log_{e}{x}$ are in product form and it can be differentiated by the product rule of differentiation.

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $\sin{x} \times \dfrac{d}{dx}{\,\log_{e}{x}}$ $+$ $\log_{e}{x} \times \dfrac{d}{dx}{\,\sin{x}}$

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $\sin{x} \times \dfrac{1}{x}$ $+$ $\log_{e}{x} \times \cos{x}$

The differentiation of the given exponential function is completed. So, we have to concentrate on simplifying the differentiated function.

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $\dfrac{\sin{x} \times 1}{x}$ $+$ $\log_{e}{x} \cos{x}$

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $\dfrac{\sin{x}}{x}$ $+$ $\log_{e}{x} \cos{x}$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $y \times \Big(\dfrac{\sin{x}}{x}$ $+$ $\log_{e}{x} \cos{x}\Big)$

The value of the function $y$ is already given in this problem. So, the variable $y$ can be replaced by its equivalent function.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $x^{\,\displaystyle \sin{x}} \times \Big(\dfrac{\sin{x}}{x}$ $+$ $\log_{e}{x} \cos{x}\Big)$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $x^{\,\displaystyle \sin{x}}\Big(\dfrac{\sin{x}}{x}$ $+$ $\log_{e}{x} \cos{x}\Big)$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $x^{\,\displaystyle \sin{x}}\Big(\dfrac{\sin{x}+x \times \log_{e}{x} \cos{x}}{x}\Big)$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $x^{\,\displaystyle \sin{x}}\Big(\dfrac{\sin{x}+x\log_{e}{x}\cos{x}}{x}\Big)$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{x^{\,\displaystyle \sin{x}}}{x} \Big(\sin{x}+x\log_{e}{x}\cos{x}\Big)$

The quotient rule of exponents with same base can be used for simplifying the right hand side expression in this equation.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $x^{\,\displaystyle \sin{x}-1} \Big(\sin{x}+x\log_{e}{x}\cos{x}\Big)$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dx}$ $\,=\,$ $x^{\,\displaystyle \sin{x}-1} \Big(\sin{x}+x\ln{x}\cos{x}\Big)$

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