Using First Principal Prove the derivative ?

Question

$x^{2} sin x$ $x^2sinx$ = $x (x cos x + 2 sin x)$ $x(xcosx+2sinx)$

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Answer 1 · 2018-06-20T09:44:09

Please see below.

Here.

$f (x) = x^{2} sin x \Rightarrow f (t) = t^{2} sin t$ $f(x)=x^2sinx =>f(t)=t^2sint$

We know that,

$f'(x)=lim_(t tox)(f(t)-f(x))/(t-x)$

$=lim_(t tox)(t^2sint-x^2sinx)/(t-x)$

$=lim_(t to x)(t^2sintcolor(red)(-x^2sint+x^2sint)-x^2sinx)/(t-x)...to(1)$

$=lim_(t tox)(sint(t^2-x^2)+x^2(sint-sinx))/(t-x)$

$=lim_(t tox){(sint(t^2-x^2))/(t-x)+(x^2(sint-sinx))/(t-x)}$

= $lim_(t tox){(sint(t-x)(t+x))/((t-x))+(x^2 *2cos((t+x)/2)sin((t- x)/2))/(t-x)}$

$=sinx(x+x)+2x^2cos((x+x)/2)lim_((t-x)/2 to0){sin((t-x)/2)/(2((t- x)/2))}$

$=2xsinx+2x^2cosx(1/2),...to[because lim_(theta to0) sintheta/theta=1$

$=2xsinx+x^2cosx$

$=x^2cosx+2xsinx$

$=x(xcosx+2sinx)$

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Note:

We can take $tocolor(red)(-t^2sinx+t^2sinx$ into $(1)$

in place of $tocolor(red)(-x^2sint+x^2sint)$