y-sinx-cos-3-x-find-y-

Question Number 204054 by hardmath last updated on 05/Feb/24

y = \sqrt{sinx} + \cos^{3} x

find : y^{^{'}} = ?

Answered by AST last updated on 05/Feb/24

y^{'} = \frac{c o s (x)}{2 \sqrt{s i n x}} - 3 s i n (x) {c o s}^{2} x

Answered by Mathspace last updated on 06/Feb/24

y^{^{'}} = \frac{{(s i n x)}^{^{'}}}{2 \sqrt{s i n x}} + 3 {c o s}^{2} x {(c o s x)}^{'}

= \frac{c o s x}{2 \sqrt{s i n x}} - 3 s i n x {c o s}^{2} x