Question

Find the value of \[\dfrac{{dy}}{{dx}}\], where \[y\] is given by\[y = {x^{sinx}} + \sqrt x \], at the point \[x = \dfrac{\pi }{2}\],
A. \[\;{\bf{1}} + \dfrac{1}{{\sqrt {2\pi } }}\]
B. \[{\bf{1}}\]
C. \[\dfrac{1}{{\sqrt {2\pi } }}\]
D. \[\;{\bf{1}} - \dfrac{1}{{\sqrt {2\pi } }}\]

Answer 1

Hint: Firstly we will differentiate \[{x^{sinx}}\]and\[\sqrt x \] separately and find the value of \[\dfrac{{dy}}{{dx}}\] and lastly we will find the value of the point \[x = \dfrac{\pi }{2}\].

Formula used:
The following derivative formula will be useful for this question
 \[\begin{array}{l}\;\;\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\\\log {x^b} = b\log x\\\sin \dfrac{\pi }{2} = 1\\\cos \dfrac{\pi }{2} = 0\end{array}\]

Complete step-by-step answer:
We know the given equation is,
\[y = {x^{sinx}} + \sqrt x \]
Let us consider,
\[u = {x^{sinx}}\]
\[v = \sqrt x \]
Firstly, we will differentiate \[u = {x^{sinx}}\] with respect to \[x\]

Taking \[\log \] both side
\[\begin{array}{l}\log u = \log {x^{sinx}}\\\log u = \sin x\log x\end{array}\]
Now, we will differentiate both sides with respect to \[x\]
\[\begin{array}{l}\dfrac{{d\log u}}{{dx}} = \dfrac{{d\sin x\log x}}{{dx}}\\\dfrac{1}{u}\dfrac{{du}}{{dx}} = \dfrac{{\sin x}}{x} + \log x\cos x\\\dfrac{{du}}{{dx}} = u\left[ {\dfrac{{\sin x}}{x} + \log x\cos x} \right]\\\dfrac{{du}}{{dx}} = {x^{sinx}}\left[ {\dfrac{{\sin x}}{x} + \log x\cos x} \right]\end{array}\]
Now, we will differentiate \[v = \sqrt x \] with respect to \[x\]
\[\dfrac{{dv}}{{dx}} = \dfrac{1}{{2\sqrt x }}\]
Now, we will differentiate the equation \[y = u + v\] with respect to \[x\]
\[\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}}\]
Putting the value of \[\dfrac{{du}}{{dx}}\] and \[\dfrac{{dv}}{{dx}}\] in the equation \[\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}}\]
\[\dfrac{{dy}}{{dx}} = {x^{sinx}}\left[ {\dfrac{{\sin x}}{x} + \log x\cos x} \right] + \dfrac{1}{{2\sqrt x }}\]
Now, we will find the value of \[\dfrac{{dy}}{{dx}}\] at the point \[x = \dfrac{\pi }{2}\]
\[\begin{array}{l}\dfrac{{dy}}{{dx}} = \dfrac{\pi }{2}\left[ {\dfrac{2}{\pi } + 0} \right] + \dfrac{1}{2}\sqrt {\dfrac{2}{\pi }} \\\dfrac{{dy}}{{dx}} = \dfrac{\pi }{2}\left[ {\dfrac{2}{\pi }} \right] + \sqrt {\dfrac{1}{{2\pi }}} \\\dfrac{{dy}}{{dx}} = 1 + \sqrt {\dfrac{1}{{2\pi }}} \end{array}\]
 Hence, the value of \[\dfrac{{dy}}{{dx}}\] is \[\;{\bf{1}} + \dfrac{1}{{\sqrt {2\pi } }}\]

Therefore, the option (A) is correct.

Additional Information: A technique for determining a function's derivative is differentiation. Mathematicians use a procedure called differentiation to determine a function's instantaneous rate of change based on one of its variables. The most popular example is the velocity or rate of change of displacement with respect to time.

Note: Always try to differentiate the given equation separately in these types of questions. The differentiation and putting the value of the function like square root, logarithm function, trigonometric function, etc should be done carefully so that it does affect the solution. Students generally make the mistake of finding the derivative of the function by having the function as their exponential power. To avoid this, the logarithm function and differentiation should be clear.