# Derivative of xsinx by First Principle

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.

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## FAQs

Q1: What is the derivative of xsinx?

Answer: The derivative of xsinx is equal to sin x+xcos x.