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

$$\mathrm{y}=\sqrt{\mathrm{sinx}}+{\cos }^3\mathrm{x}$$$$\mathrm{find}:{\mathrm{y}}^{{}^{\prime }}=?$$

Answered by deleteduser1 last updated on 05/Feb/24

$$y^{\prime }=\frac{cos\left(x\right)}{2\sqrt{sinx}}-3sin\left(x\right){cos}^{2}x$$

Answered by Mathspace last updated on 06/Feb/24

$$y^{{}^{\prime }}=\frac{{\left(sinx\right)}^{{}^{\prime }}}{2\sqrt{sinx}}+3{cos}^{2}x{\left(cosx\right)}^{\prime }$$$$=\frac{cosx}{2\sqrt{sinx}}-3sinx{cos}^{2}x$$

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