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Question: Find \(\dfrac{{dy}}{{dx}}\), when \(y = {x^x} - {2^{\sin x}}\)....

Find dydx\dfrac{{dy}}{{dx}}, when y=xx−2sin⁡xy = {x^x} - {2^{\sin x}}.

Explanation

Solution

Let u=xxu = {x^x} and v=2sin⁡xv = {2^{\sin x}}. Take logarithm of u and v. Compute dudx\dfrac{{du}}{{dx}} and dvdx\dfrac{{dv}}{{dx}}.
Then use dydx=dudx−dvdx\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} - \dfrac{{dv}}{{dx}} to get the answer.

Complete step by step solution:
Given that y=xx−2sin⁡xy = {x^x} - {2^{\sin x}}
We have to find the derivative of y with respect to x.
Letu=xxu = {x^x} and v=2sin⁡xv = {2^{\sin x}}
Then y=u−vy = u - v
We will first differentiate u and v with respect to x.
Now consider u=xxu = {x^x}.
Taking log on both the sides,
log⁡u=log⁡xx ⇒log⁡u=xlog⁡x.......(1)  \log u = \log {x^x} \\\ \Rightarrow \log u = x\log x.......(1) \\\
Here we used the property of logarithm, log⁡ab=blog⁡a\log {a^b} = b\log a
Differentiate (1) with respect to x on both the sides
d(log⁡u)dx=d(xlog⁡x)dx ⇒1ududx=(x×1x+log⁡x×1)=1+log⁡x ⇒dudx=u(1+log⁡x)  \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.
dudx=xx(1+log⁡x).....(A)\dfrac{{du}}{{dx}} = {x^x}(1 + \log x).....(A)
Consider v=2sin⁡xv = {2^{\sin x}}
Taking log on both the sides,

log⁡v=log⁡(2sin⁡x) ⇒log⁡v=sin⁡x(log⁡2)......(2) \log v = \log ({2^{\sin x}}) \\\ \Rightarrow \log v = \sin x(\log 2)......(2) \\\

Here log⁡2\log 2is a constant.
Differentiate (2) with respect to x on both the sides

d(log⁡v)dx=d(sin⁡x(log⁡2))dx ⇒1vdvdx=log⁡2d(sin⁡x)dx=log⁡2(cos⁡x) ⇒dvdx=v(log⁡2)(cos⁡x) \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.
dvdx=2sin⁡x(log⁡2)(cos⁡x)....(B)\dfrac{{dv}}{{dx}} = {2^{\sin x}}(\log 2)(\cos x)....(B)
Now, y=u−vy = u - v
Differentiate with respect to x on both the sides
dydx=dudx−dvdx\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} - \dfrac{{dv}}{{dx}}
Using (A) and (B), we get
dydx=xx(1+log⁡x)−2sin⁡x(log⁡2)(cos⁡x)\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