Question

If$y=x\sin x$, then

& A.\left( \dfrac{1}{y} \right)\left( \dfrac{dy}{dx} \right)=\dfrac{1}{x}+\cot x \\\ & B.\dfrac{dy}{dx}=\dfrac{1}{x}+\cot x \\\ & C.y\dfrac{dy}{dx}=\dfrac{1}{x}-\cot x \\\ & D.none of these \\\ \end{aligned}$$

Answer 1

In order to solve $y=x\sin x$, we will be differentiating each term of $y=x\sin x$ using the product rule. After differentiating it, we will be solving for each term. Upon solving it, we obtain the required solution.

Complete step-by-step solution:
Let us have a brief regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between ${{0}^{\circ }}$ and ${{360}^{\circ }}$. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us find the value of $y=x\sin x$
Upon differentiating it, we get
$\dfrac{dy}{dx}y=\dfrac{dy}{dx}x\sin x$
We will be differentiating it using the product rule i.e. ${{\left( uv \right)}^{'}}={{u}^{'}}v+uv'$
So solving accordingly, we get
$\dfrac{dy}{dx}=x\cos x+\sin x$
$\Rightarrow \dfrac{1}{y}\left( \dfrac{dy}{dx} \right)=\dfrac{x\cos x}{x\sin x}+\dfrac{\sin x}{x\sin x}$
Upon simplifying it, we get
$\Rightarrow \cot x+\dfrac{1}{x}$
\therefore $$$$y=x\sin x$$$$=\cot x+\dfrac{1}{x}
Hence option A is the correct answer.

Note: We must always try to expand the functions into the identities whose values we are aware of, for easy calculations. While differentiating the functions we must be aware of the multiplication rule, division rule, the addition rule and the chain rule. We must check for the rule that relates and try to convert into that particular form for simplification. In the above problem we used the method of multiplication which is also called the product rule.