Question

Evaluate \[\dfrac{d}{dx}\sqrt{x\,\sin x}\].
a) $\dfrac{(\sin x+x\cos x)}{2\sqrt{(x\sin x)}}$
b) $\dfrac{(\sin x+x\cos x)}{\sqrt{(x\sin x)}}$
c) $\dfrac{(x\sin x+x\cos x)}{2\sqrt{(x\sin x)}}$
d) $\dfrac{(x\sin x+x\cos x)}{\sqrt{(2x\sin x)}}$

Answer 1

Hint: We need to find the differentiation of \[\sqrt{x\,\sin x}\] with respect to x. First we need to apply chain rule, that is $\dfrac{d}{d x}[f(g(x))]=f^{\prime}(g(x)) g^{\prime}(x)$. After this we apply product rule for further simplification , that is the product rule is$\dfrac{d}{d x}[f(x) g(x)]=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$.

Formula used:
$\dfrac{d}{d x}[f(x) g(x)]=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$.

Complete step by step solution:
Let \[y=\sqrt{x\,\sin x}\]
Differentiate with respect to x.
 \[\dfrac{dy}{dx}=\dfrac{d}{dx}\sqrt{x\sin x}\]
We know that \[\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}\], then
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x\sin x}}\dfrac{d}{dx}\left( x\sin x \right)\,\,\,--(1)\]
Now to find \[\dfrac{d}{dx}\left( x\sin x \right)\] we need to apply product rule,
\[\dfrac{d}{dx}\left( x\sin x \right)=x\dfrac{d}{dx}\sin x+\sin x\dfrac{d}{dx}x\]
We know that differentiation of sine with respect to x is cosine.
\[\dfrac{d}{dx}\left( x\sin x \right)=x.\cos x+\sin x.1\]
\[\Rightarrow \dfrac{d}{dx}\left( x\sin x \right)=x.\cos x+\sin x\]
Substituting this in equation 1 we have,
\[\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x\sin x}}\left( x.\cos x+\sin x \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{(\sin x+x\cos x)}{2\sqrt{(x\sin x)}}\]

Hence, option a) is correct.

Note: In order to solve this problem we need to know the differentiation of all trigonometric functions. We also have quotient rule, that is $\dfrac{d}{d x}\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{[g(x)]^{2}}$. According to the given problem we apply the differentiation rules. We also know that \[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\].