Question

Find ${\displaystyle \frac{dy}{dx}}$, when $y=x^x-2^{\sin x}$.

Answer 1

Hint: Let $u=x^x$ and $v=2^{\sin x}$. Take logarithm of u and v. Compute $${\displaystyle \frac{du}{dx}}$$ and $${\displaystyle \frac{dv}{dx}}$$.
Then use ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{du}{dx}}-{\displaystyle \frac{dv}{dx}}$ 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,
$$\begin{gathered} \log u=\log x^x \\ \Rightarrow \log u=x\log x.......(1) \end{gathered}$$
Here we used the property of logarithm, $\log a^b=b\log a$
Differentiate (1) with respect to x on both the sides
$$\begin{gathered} {\displaystyle \frac{d(\log u)}{dx}}={\displaystyle \frac{d(x\log x)}{dx}} \\ \Rightarrow {\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}=(x\times {\displaystyle \frac{1}{x}}+\log x\times 1)=1+\log x \\ \Rightarrow {\displaystyle \frac{du}{dx}}=u(1+\log x) \end{gathered}$$
Now, substitute the value of u.
${\displaystyle \frac{du}{dx}}=x^x(1+\log x).....(A)$
Consider $v=2^{\sin x}$
Taking log on both the sides,
$$\begin{gathered} \log v=\log (2^{\sin x}) \\ \Rightarrow \log v=\sin x(\log 2)......(2) \end{gathered}$$
Here $\log 2$is a constant.
Differentiate (2) with respect to x on both the sides
$$\begin{gathered} {\displaystyle \frac{d(\log v)}{dx}}={\displaystyle \frac{d(\sin x(\log 2))}{dx}} \\ \Rightarrow {\displaystyle \frac{1}{v}}{\displaystyle \frac{dv}{dx}}=\log 2{\displaystyle \frac{d(\sin x)}{dx}}=\log 2(\cos x) \\ \Rightarrow {\displaystyle \frac{dv}{dx}}=v(\log 2)(\cos x) \end{gathered}$$
Now, substitute the value of v.
$${\displaystyle \frac{dv}{dx}}=2^{\sin x}(\log 2)(\cos x)....(B)$$
Now, $y=u-v$
Differentiate with respect to x on both the sides
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{du}{dx}}-{\displaystyle \frac{dv}{dx}}$
Using (A) and (B), we get
${\displaystyle \frac{dy}{dx}}=x^x(1+\log x)-2^{\sin x}(\log 2)(\cos x)$ 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