Question

Differentiation -

y = sin√x ÷ √sinx

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QUALITY ANSWER NEEDED

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Answer 1

Given :

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

Apply the quotient rule :

$\implies \dfrac{d}{dx}[\dfrac{sin\sqrt{x}}{\sqrt{sinx}}]\\\\\implies [\dfrac{\dfrac{d}{dx}sin\sqrt{x}.\sqrt{sinx}-\dfrac{d}{dx}(\sqrt{sinx})sin\sqrt{x}}{(\sqrt{sinx})^2}]\\\\\implies \dfrac{cos\sqrt{x}\times \dfrac{1}{2}sinx^{1/2-1}-\dfrac{1}{2}sinx^{1/2-1}\times\sqrt{cosx}\times sin\sqrt{x}}{sinx}\\\\\implies \dfrac{\dfrac{\cos\left(\sqrt{x}\right)\sqrt{\sin\left(x\right)}}{2\sqrt{x}}-\dfrac{\sin\left(\sqrt{x}\right)\cos\left(x\right)}{2\sqrt{\sin\left(x\right)}}}{\sin\left(x\right)}$

That seems an awkward answer but after class 10 , awkward things do happen .

Answer 2

Hey dear ‼️

plzz refer to the attached file......

keep loving...❤️❤️❤️❤️❤️❤️✌️

have uh more such type of question?