Question

Find the derivative of `sin^2x` using first principles?

Answer 1

See below

Explanation:

It is tedious (and space-consuming) to use `lim_(hrarr0)` on every line, so I hope you'll excuse my approach. We'll simplify the difference quotient first, then find the limit.

`((sin(x+h))^2-(sinx)^2)/h = ((sinxcos h +cos x sin h)^2-(sinx)^2)/h`

`= (color(red)(sin^2xcos^2h)+2sinxcos h cosx sin h +color(blue)(cos^2xsin^2h)-color(red)(sin^2x))/h`

`= (sin^2x(cos^2h-1))/h +(2sinxcos h cosx sin h)/h +(cos^2xsin^2h)/h`

`= sin^2x(cosh-1)/h(cosh+1) +2sinxcos h cosx sin h /h+cos^2xsin h/h sin h`

Taking limit as `h rarr0`, we get

`sin^2(0)(0)(cos 0 +1) +2sinx (cos 0) cosx (1) +cos^0 (1) sin0`

`= 0+2sinx cos x +0`

`= 2sinx cos x`

Answer 2

`f'(x) =2sinxcosx`

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`