Question

Differentiate the function w.r.t. x
$$x^x-2^{\sin x}$$

Answer 1

Hint: Suppose the given value $$\left(x^x-2^{\sin x}\right)$$ into two variables. Thereafter, we will solve separately, by using differentiation of the function with respect to $$\text{x}$$.

Complete step by step solution:
Let$$y=x^x-2^{\sin x}$$
Also, let $$x^x=u$$ and $$2^{\sin x}=v$$
$$\therefore y=u-v$$
Differentiating both sides with respect to$$\text{x}$$.
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{du}{dx}}-{\displaystyle \frac{dv}{dx}}$$
First we will solve: $$u=x^x$$
Taking logarithm on both sides, we obtain
$$\log u=xlogx$$
Differentiating both sides with respect to$$\text{x}$$, we obtain
$${\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}=\left[\log x\times {\displaystyle \frac{d}{dx}}\left(x\right)+x\times {\displaystyle \frac{d}{dx}}\left(\log x\right)\right]$$
$$\Rightarrow {\displaystyle \frac{du}{dx}}=u\left[\log x\times 1+x\times {\displaystyle \frac{1}{x}}{\displaystyle \frac{d}{dx}}\left(x\right)\right]$$
$${\displaystyle \frac{du}{dx}}=x^2\left(\log x+1\right)$$ $$\left(\because u=x^2\right)$$
$$v=2^{\sin x}$$
Taking logarithm on both the sides with respect to $$\text{x}$$, obtain
$$\log v=sinxlog2$$
Differentiating both sides with respect to x we obtain
$${\displaystyle \frac{1}{v}}.{\displaystyle \frac{dv}{dx}}=\log 2.{\displaystyle \frac{d}{dx}}\sin x$$ $$\left(\therefore \log 2\right)$$is a constant term
$$\Rightarrow {\displaystyle \frac{dv}{dx}}=v\log 2\cos x$$
$$\Rightarrow {\displaystyle \frac{dv}{dx}}=2^{\sin x}\cos x\log 2$$
Therefore, adding the values of ${\displaystyle \frac{du}{dx}}and{\displaystyle \frac{dv}{dx}}$, we will get
$$\therefore {\displaystyle \frac{dy}{dx}}=x^x\left(1+\log x\right)-2^{\sin x}\cos x\log 2$$

Note: To differentiate something means to take the derivative of that value. Taking the derivative of a function is the same as finding the slope at any point, so differentiating is just finding the slope.