Question

Find the derivative of xsinx using first principle?

Answer 1

if y=f(x) g (x)
then differentiation w.r.t.x
dy/dx=f (x) dg(x)/dx+g(x)df(x)/dx
now use this
y=x.sinx
dy/dx=x.d (sinx)/dx+sinx.dx/dx
=x.cosx+sinx

Answer 2

Let $f(x)=x\sin x$ . Then the derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{(x+h)\sin(x+h)-x\sin x}hf </p><p>′$

Expand the numerator:

$(x+h)\sin(x+h)-x\sin x=x(\sin x\cos h+\sin h\cos x)+h(\sin x\cos h+\sin h\cos x)-x\sin x(x+h)sin(x+h)−$

This can be rewritten as

$x\sin x(\cos h-1)+x\cos x\sin h+h(\sin x\cos h+\sin h\cos x)$

So you're left with the limit

$\displaystyle\lim_{h\to0}\frac{x\sin x(\cos h-1)+x\cos x\sin h+h(\sin x\cos h+\sin h\cos x)}h$

which you can split up as a sum of factored limits

$\displaystyle x\sin x\lim_{h\to0}\frac{\cos h-1}h+x\cos x\lim_{h\to0}\frac{\sin h}h+\lim_{h\to0}\frac{h(\sin x\cos h+\sin h\cos x)}$

The first two limits use two special limits,

$\displaystyle\lim_{h\to0}\frac{1-\cos h}h=0\quad\text{and}\quad\lim_{h\to0}\frac{\sin h}$

while for the last, you can cancel out the factors of hh in the numerator and denominator. You're left with

$\displaystyle 0+x\cos x+\lim_{h\to0}(\sin x\cos h+\sin h\cos x)0+xcosx+ </p><p>h→0</p><p>lim</p><p> </p><p> (sinxcosh+sinhcosx)</p><p>You have \lim\limits_{h\to0}\cos h=1 h→0lim cosh=1 and \lim\limits_{h\to0}\sin h=0 h→0lim sinh=0 , so the derivative is</p><p>f'(x)=x\cos x+\sin xf ′ (x)=xcosx+sinx$