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**Hint:**Let $u = {x^x}$ and $v = {2^{\sin x}}$. Take logarithm of u and v. Compute \[\dfrac{{du}}{{dx}}\] and \[\dfrac{{dv}}{{dx}}\].

Then use $\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} - \dfrac{{dv}}{{dx}}$ to get the answer.

**Complete step by step solution:**

Given that $y = {x^x} - {2^{\sin x}}$

We have to find the derivative of y with respect to x.

Let$u = {x^x}$ and $v = {2^{\sin x}}$

Then $y = u - v$

We will first differentiate u and v with respect to x.

Now consider $u = {x^x}$.

Taking log on both the sides,

$

\log u = \log {x^x} \\

\Rightarrow \log u = x\log x.......(1) \\

$

Here we used the property of logarithm, $\log {a^b} = b\log a$

Differentiate (1) with respect to x on both the sides

$

\dfrac{{d(\log u)}}{{dx}} = \dfrac{{d(x\log x)}}{{dx}} \\

\Rightarrow \dfrac{1}{u}\dfrac{{du}}{{dx}} = (x \times \dfrac{1}{x} + \log x \times 1) = 1 + \log x \\

\Rightarrow \dfrac{{du}}{{dx}} = u(1 + \log x) \\

$

Now, substitute the value of u.

$\dfrac{{du}}{{dx}} = {x^x}(1 + \log x).....(A)$

Consider $v = {2^{\sin x}}$

Taking log on both the sides,

\[

\log v = \log ({2^{\sin x}}) \\

\Rightarrow \log v = \sin x(\log 2)......(2) \\

\]

Here $\log 2$is a constant.

Differentiate (2) with respect to x on both the sides

\[

\dfrac{{d(\log v)}}{{dx}} = \dfrac{{d(\sin x(\log 2))}}{{dx}} \\

\Rightarrow \dfrac{1}{v}\dfrac{{dv}}{{dx}} = \log 2\dfrac{{d(\sin x)}}{{dx}} = \log 2(\cos x) \\

\Rightarrow \dfrac{{dv}}{{dx}} = v(\log 2)(\cos x) \\

\]

Now, substitute the value of v.

\[\dfrac{{dv}}{{dx}} = {2^{\sin x}}(\log 2)(\cos x)....(B)\]

Now, $y = u - v$

Differentiate with respect to x on both the sides

$\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} - \dfrac{{dv}}{{dx}}$

Using (A) and (B), we get

$\dfrac{{dy}}{{dx}} = {x^x}(1 + \log x) - {2^{\sin x}}(\log 2)(\cos x)$ which is the required answer.

**Note:**Whenever we have two functions given like in this question,make sure to find the derivative of each of the function and separately and then add/subtract both the derivatives.Do not solve it directly

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