Find the derivative of the following functions using first principle of derivative.
(i) sin (3x + 2) (ii) `"sin" (x^(2) + 5)` (iii) x cos x

Explanation

(i) Let `f(x) = sin (3x + 2)` Then `f(x + h) = sin (3(x + h) + 2)`
`:.f'(x) = underset(h to 0)(lim) (sin(3(x + h) + 2) - sin (3x + 2))/(h)`
`= underset(h to 0)(lim) (2 cos ((3x + 3h + 2 + 3x + 2)/(2)) sin ((3x + 3h + 2 - 3x - 2)/(2)))/(h)`
`= underset(h to 0)(lim) cos ((6x + 3h + 4)/(2)) underset(h to 0)(lim) ("sin"(3h)/(2) 3)/((3h)/(2))`
`= 3 cos (3x +2)`
`:.f(sin(3x + 2)) = 3 cos (3x + 2)`
(ii) `f(x = sin(x^(2) + 5)`. Then `f(x + h) = sin ((x + h)^(2) + 5)`
`:.f'(x) = underset(h to 0)(lim)(sin (x + h)^(2) + 5 - sin (x^(2) + 5))/(h)`
`underset(h to 0)(lim) (2 cos (((x + h)^(2) + 5 + x^(2) + 5)/(2)) sin (((x + h)^(2) + 5 - (x^(2) + 5))/(2)))/(h)`
`= undeset(h to 0) 2 cos (2x^(2) + h^(2) + 2xh + 10)/(2)) (sin((h^(2) + 2xh)/(2)))/(h)`
`= underset(h to 0)(lim) 2 cos (2x^(2) + h^(2) + 2xh + 10)/(2)) underset(h to 0)(lim) (sin((h^(2) + 2xh)/(2)))/(h(h + 2x))/(2)) ((h + 2x)/(2))`
`= 2 cos (x^(2) + 5) .1.x = 2x cos (x^(2) + 5)`
`:.f'(sin(x^(2) + 5)) - 2x cos (x^(2) + 5)`
(iii) Let `f(x) = x cos x`. then `f(x + h) = (x + h) cos (x + h)`
`:.f'(x) = underset(h to 0)(lim) ((x + h) cos (x + h) - x cos x)/(h)`
`= underset(h to 0)(lim) (x cos (x + h) - x cos x + h cos (x + h))/(h)`
`= underset(x to 0)(lim) (x[cox(x + h)- cos x])/(h) + underset(h to 0)(lim) (h cos (x + h))/(h)`
`= x underset(h to 0)(lim) (-sin((2x + h)/(2)) sin ((h)/(2)))/((h)/(2)) + underset(h to 0)(lim) cos (x + h)`
`= x (-sin x) + cos x = cosx = x sin x`
`:.f'(x cos x) = cos x - x sin x`