Question

Find the derivatives w.r.t.x from first principle:
x sin x

Answer 1

❤what we have to do with this value XSINX

Answer 2

Answer:

Derivative of x.sinx = x.cosx + sinx

Step-by-step explanation:

In the question,

We have to calculate the derivative of xsinx from the first principle,

We know that first principle states that,

$f'(x)= \lim_{h \to \ 0} \frac{f(x+h)-f(x)}{h}$

So,

$f(x)=x.sinx$

Therefore,

$f(x+h)=(x+h).sin(x+h)$

So,

On putting we get,

$f'(x)= \lim_{h \to \ 0} \frac{(x+h)sin(x+h)-xsinx}{h}=\frac{x[sin(x+h)-sinx]+hsin(x+h)}{h}\\ =\frac{x(2cos(\frac{2x+h}{2}).sin\frac{h}{2})}{h}+\frac{hsin(x+h)}{h}\\ =\frac{x(2cos(\frac{2x+h}{2})).(sin\frac{h}{2})}{h/2}+\frac{hsin(x+h)}{h}\\= \lim_{h \to \ 0}\frac{x(cos(\frac{2x+h}{2}))}{1}\times \frac{sin(h/2)}{(h/2)}+ \lim_{h \to \ 0}sin(x+h)\\=x.cosx+sinx$

Therefore, on putting the value of h = 0 we get the required derivative.

So,

Derivative of x.sinx = x.cosx + sinx