Question

Differentiation -

y = sin√x + √sinx

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QUALITY ANSWER NEEDED​

Answer 1

Given :

$y=sin\sqrt{x}+\sqrt{sinx}$

$\implies \dfrac{dy}{dx}=\dfrac{d}{dx}[sin\sqrt{x}+\sqrt{sinx}]\\\\\implies \dfrac{d}{dx}[(sinx)^{1/2}+\sqrt{sinx}]$

Apply chain rule now and proceed :

$\implies \dfrac{d}{dx}[\dfrac{1}{2}sin^{1/2-1}x.\dfrac{d}{dx}sinx+cos\sqrt{x}.\dfrac{d}{dx}\sqrt{x}]$

$\implies \dfrac{1}{2}sin^{-1/2}x.cosx+cos\sqrt{x}(\dfrac{1}{2}x^{1/2-1})\\\\\implies \dfrac{1}{2sin\sqrt{x}}.cosx+cos\sqrt{x}\times \dfrac{1}{2\sqrt{x}}\\\\\implies \dfrac{cosx}{2sin\sqrt{x}}+\dfrac{cos\sqrt{x}}{2\sqrt{x}}\\\\\implies \dfrac{cos\sqrt{x}}{2}(\dfrac{cos\sqrt{x}}{sin\sqrt{x}}+\dfrac{1}{\sqrt{x}})\\\\\implies \boxed{\dfrac{cos\sqrt{x}}{2}(cot\sqrt{x}+\dfrac{1}{\sqrt{x}})}$

Answer 2

Hey dear ‼️

plzz refer to the attached file..

keep loving..❤️❤️❤️❤️❤️✌️