Find the derivative of the following from the first principle .
(i) `sin^(2)x`, (ii) `cos^(2)x`

Explanation

(i) Let `y = sin^(2)x`.
Let `deltay` be an increment in y, corresponding to an increment `deltax` in x.
Then, `y + deltay = sin^(2) (x+deltax)`
`rArr deltay = sin^(2) (x+deltax) - sin^(2)x`
`rArr (deltay)/(deltax) = (sin^(2)(x+deltax) - sin^(2)x)/(deltax)`
`rArr (dy)/(dx) = underset(deltaxrarr0)("lim") (deltay)/(deltax)`
`= underset(deltaxrarr0)("lim") (sin^(2)(x+deltax) - sin^(2)x)/(deltax)`
`=underset(deltaxrarr0)("lim") (sin(2x+deltax) sindeltax)/(deltax)`
`[:' sin^(2)A - sin^(2)B = sin(A+B).sin(A-B)]`
`=underset(deltaxrarr0)("lim") sin(2x+deltax).underset(deltaxrarr0)("lim")(sindeltax)/(deltax)`
`= (sin2x xx 1)=sin2x`.
Hence, `d/(dx) (sin^(2)x) = sin 2x`.
Let `y = cos^(2)x` .
Let `deltay` be an increment in y, corresponding to an increment `deltax` in x.
Then, `y + deltay = cos^(2)(x+deltax)`.
`rArr deltay = cos^(2) (x+deltax) - cos^(2)x`
`rArr (deltay)/(deltax) = (cos^(2)(x+deltax)-cos^(2)x)/(deltax)`
`rArr (dy)/(dx)= underset(deltaxrarr0)("lim") (deltay)/(deltax)`
`= underset(deltaxrarr0) ("lim") (cos^(2)(x+deltax) - cos^(2)x)/(deltax)`
`= underset(deltararr0) ("lim") ([cos(x+deltax)+cosx] xx [cos(x+deltax) - cosx])/(deltax)`
`= underset(deltaxrarr0)("lim") ({2cos(x+(deltax)/(2)) cos '((deltax)/(2))}{-2sin(x+(deltax)/(2))sin ((deltax)/(2))})/(deltax)`
`= - 2 underset(deltaxrarr0)("lim")cos(x+(deltax)/(2)). underset(deltaxrarr0)("lim") cos((deltax)/(2)).underset(deltaxrarr0)("lim") sin(x+(deltax)/(2)). underset(deltaxrarr0)("lim") (sin(deltax//2))/((deltax//2))`
`= (-2cos x xx 1 xx sin x xx 1) = - 2 sin x cos x = - sin 2x`.
Hence, `d/(dx) (cos^(2x)x) = - sin 2x`.