Question

How do you find the differential $dy$ of the function $y=x\sin x$ ?

Answer 1

From the question given we have the function $y=x\sin x$ and now we have to find the differential $dy$of the function. Now we have to differentiate both sides with respect to the x then we have to use the uv rule in the function while differentiating the left hand side part. UV rule means when we differentiate any function in the form UV then we will have to differentiate like $D\left( UV \right)=D\left( U \right)\times V+D\left( V \right)\times U$.

Complete step by step solution:
From the question, given function is
$\Rightarrow y=x\sin x$
Now, we have to differentiate both sides with respect to the x.
By differentiating both sides with respect to x we will get,
$\Rightarrow y=x\sin x$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( x\sin x \right)$
Here the left-hand side part $x\sin x$ is in the form of UV.
We know that if any function is in the form of UV then we will use UV rule for differentiating the function.
UV rule means,
$\Rightarrow D\left( UV \right)=D\left( U \right)\times V+D\left( V \right)\times U$
here U is x and V is $\sin x$.
Then by substituting in the above rule we will get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{dx}{dx}\times \sin x+\dfrac{d\left( \sin x \right)}{dx}\times x$
We know that differentiation of x with respect to x is $1$
$\Rightarrow \dfrac{dy}{dx}=1\times \sin x+\dfrac{d\left( \sin x \right)}{dx}\times x$
We know that differentiation of $\sin x$ with respect to x is $\cos x$.
Then we will get,
$\Rightarrow \dfrac{dy}{dx}=1\times \sin x+\cos x\times x$
$\Rightarrow \dfrac{dy}{dx}=\sin x+x\cos x$
Therefore, we want the differential $dy$. So, shift the $dx$ from left hand side to the right-hand side we will get,
$\Rightarrow dy=\left( \sin x+x\cos x \right)dx$

Note:
Students should know the basic differentiation formulas like,

& \Rightarrow \dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{n-1}} \\\ & \Rightarrow \dfrac{d\left( \sin x \right)}{dx}=\cos x \\\ & \Rightarrow \dfrac{d\left( \cos x \right)}{dx}=-\sin x \\\ & \Rightarrow \dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x \\\ & \Rightarrow \dfrac{d\left( \cot x \right)}{dx}=-\cos e{{c}^{2}}x \\\ & \Rightarrow \dfrac{d\left( \sec x \right)}{dx}=\sec x\times \tan x \\\ & \Rightarrow \dfrac{d\left( \cos ecx \right)}{dx}=-\cos ecx\times \cot x \\\ & \Rightarrow \dfrac{d\left( cons\tan t \right)}{dx}=0 \\\ \end{aligned}$$ Students should also know the UV rule. $$\Rightarrow D\left( UV \right)=D\left( U \right)\times V+D\left( V \right)\times U$$ student should also know about the U by V rule, if any function is in the form U by V then the differentiating that function will be done by U by V rule, $$\Rightarrow D\left( \dfrac{U}{V} \right)=\dfrac{V\times D\left( U \right)-U\times D\left( V \right)}{{{V}^{2}}}$$ . Students should not make any calculation mistakes.