Using First Principal Prove the derivative ?

Question

#x^2sinx# = #x(xcosx+2sinx)#

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Answer 1 · 2018-06-20T09:44:09

Please see below.

Here.

#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)#

...........................................................................................

Note:

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

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