Question

Differentiate sin^2x

from first principles​

Answer 1

AnswEr:-

From first principles,

$=> f'(x) =  \lim_{h \to 0}  \frac{f(x+h)-f(x)}{h}\\ \\ \\Given \: , f(x) = sin^{2} x \\ \\ \\=> f'(x) =  \lim_{h \to 0}   \frac{sin^2 (x+h)-sin^2 (x)}{h} \\ \\ \\=>  \frac{(sinx.cos h  + cosx.sinh)^2 - (sinx)^2}{h}  \\ \\ \\=> \frac{sin^2 x . cos^2 h  +2sinx. cosh. cosx.sinh + cos^2 x.sin^2 h - sin^2 x}{h} \\ \\ \\=> sin^2 x \:  \frac{cosh-1}{h} (cosh+1) + 2sinx.cosh.cosx.\frac{sinh}{h} + cos^2 x .sinh .\frac{sinh}{h} \\ \\ \\As \: \:  h \to 0 \:  \:  we  \: get,$

=> sin² (0)(0)(cos 0 +1 ) + 2 sinx (cos 0)cosx(1) +cos°(1)sin0

            = 0 + 2sinx cosx + 0

           = 2sinx cosx.

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Answer 2

Step-by-step explanation:

The definition of the derivative of y=f(x) is

f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h

So with f(x) = sin^2x then;

f'(x)=lim_(h rarr 0) {sin^2(x+h) -sin^2(x)}/h

Let us focus on the numerator f(x+h)-f(x);

f(x+h)-f(x)

\ \ = sin^2(x+h) -sin^2(x)

\ \ = (sinxcos h+cosxsin h)^2 -sin^2(x)

\ \ = (sinxcos h)^2+2sinxcos hcosxsin h + (cosxsin h)^2 -sin^2(x)

\ \ = sin^2xcos^2h+2sinxcos hcosxsin h + cos^2xsin^2h -sin^2x

\ \ = sin^2xcos^2h+2sinxcos hcosxsin h + cos^2xsin^2h -sin^2x

\ \ = sin^2x(cos^2h-1)+2sinxcos hcosxsin h + cos^2xsin^2h

\ \ = sin^2xsin^2h+2sinxcos hcosxsin h + cos^2xsin^2h

\ \ = sin^2h(sin^2x+cos^2x) + 2sinxcos hcosxsin h

\ \ = sin^2h+ 2sinxcos hcosxsin h

And so the limit becomes:

f'(x)=lim_(h rarr 0) {sin^2h+ 2sinxcos hcosxsin h}/h

" "=lim_(h rarr 0) {sin^2h/h+ (2sinxcos hcosxsin h)/h}

" "=lim_(h rarr 0) {sin h * sin h/h+ 2sinxcosx*cos h*sin h/h}

And then using the Fundamental trigonometric calculus limits:

lim_(theta rarr0) sin theta /theta = 1

we have:

f'(x)= 0 * 1+ 2sinxcosx*1*1

" "=2sinxcosx