The derivative of xsin x is sin x+x cos x. In this post, we will find the differentiation of x sin x using the first principle of derivatives. The derivative formula of xsin x is given below:
$\dfrac{d}{dx}(x \sin x)$ $=\sin x+x\cos x$.
We will now prove the above formula.
Derivative of xsin x by First Principle
If $f(x)$ is a function of $x$, then its derivative from first principle is determined by the following limit:
$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$
Take $f(x)=x\sin x$ in the above limit.
So the differentiation of x sin x using the first principle is
$\dfrac{d}{dx}(x\sin x)$ $=\lim\limits_{h \to 0} \dfrac{(x+h)\sin(x+h)-x\sin x}{h}$
$=\lim\limits_{h \to 0} \dfrac{x\sin(x+h)+h\sin(x+h)-x\sin x}{h}$
Rearranging the numerator, the above limit
$=\lim\limits_{h \to 0} \dfrac{x\sin(x+h)-x\sin x+h\sin(x+h)}{h}$
$=\lim\limits_{h \to 0}$ $[\dfrac{x\sin(x+h)-x\sin x}{h}+\dfrac{h\sin(x+h)}{h}]$
Now applying the addition property of limits, we get that
$=\lim\limits_{h \to 0}\dfrac{x[\sin(x+h)-\sin x]}{h}$+ $\lim\limits_{h \to 0} \frac{h\sin(x+h)}{h}$
$=\lim\limits_{h \to 0}\dfrac{x[2\cos \frac{x+h+x}{2} \sin \frac{x+h-x}{2}]}{h}$+ $\lim\limits_{h \to 0} \sin(x+h)$ as we know the formula sin a – sin b = 2 cos(a+b)/2 sin(a-b)/2.
$=\lim\limits_{h \to 0}\dfrac{2x \cos (x+h/2) \sin h/2}{h}$+ $\lim\limits_{h \to 0} \sin(x+h)$
$=x \cos (x+0/2) \lim\limits_{h \to 0}\dfrac{\sin \frac{h}{2}}{h/2}$+ $\lim\limits_{h \to 0} \sin(x+0)$
[Let h/2 = t. Then t tends to 0 as h tends \to 0.]
$=x \cos x \lim\limits_{t \to 0}\dfrac{\sin t}{t}$+ $\sin x$
$=x \cos x \cdot 1$+ $\sin x$ as the limit of (sin x)/x is 1 when x tends to zero.
$=x\cos x+\sin x$
So the derivative of x sin x is xcosx+sin x, and this is obtained by the first principle of derivatives.
Also Read:
Derivative of Fourth Root of x
FAQs
Q1: What is the derivative of xsinx?
Answer: The derivative of xsinx is equal to sin x+xcos x.