Question

differentiate with respect to X☝️​

Answer 1

Answer:

$\frac{\sqrt{1+cosx} }{2}$

Step-by-step explanation:

Before Differentiating, let us multiply numerator and denominator with

$\sqrt{1 - cosx}$ as there would be no change

So, now the equation turns,

$\frac{\sin x}{\sqrt{1 + cosx}} = \frac{sinx}{\sqrt{1+cosx}} \times \frac{\sqrt{1 - cosx}}{\sqrt{1-cosx}}$

$=> \frac{sinx \times \sqrt{1-cosx}}{\sqrt{1 - cos^2x}}$

$=> \frac{sinx \times \sqrt{1 - cosx}}{\sqrt{sin^2x}} ---- [1 - cos^2x = sin^2x]$

$=> \frac{sinx \times \sqrt{1 - cosx}}{sinx}$

$=> \sqrt{1 - cosx}$

Now, differentiating √(1 - cosx) , we get,

$\frac{d}{dx}(\sqrt{1 - cosx} ) = \frac{1}{2\sqrt{1 -cosx} } \times \frac{d}{dx}(-cosx) \\\\=> \frac{sinx}{2\sqrt{1 - cosx} }\ or\ \frac{\sqrt{1+cosx} }{2}$

Hope it helps

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