Differentiate the following with respect to x:

Question:

Differentiate the following with respect to x:

$\sqrt{\cot \sqrt{x}}$

 

Solution:

To Find: Differentiation

NOTE : When 2 functions are in the product then we used product rule i.e

$\frac{d(u, v)}{d x}=v \frac{d u}{d x}+u \frac{d v}{d x}$

Formula used: $\frac{d}{d x}(\sqrt{\cot \sqrt{x}})=\frac{1}{2 \sqrt{\cot \sqrt{x}}} \times \frac{d}{d x}(\cot \sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x})$

Let us take $y=\sqrt{\cot \sqrt{x}}$

So, by using the above formula, we have

$\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{\cot \sqrt{\mathrm{x}}}=\frac{1}{2 \sqrt{\cot \sqrt{\mathrm{x}}}} \times \frac{\mathrm{d}}{\mathrm{dx}} \cot$

$\sqrt{x} \times \frac{d}{d x} \sqrt{x}=\frac{1}{2 \sqrt{\cot \sqrt{x}}} \times\left(-\sec ^{2} \sqrt{x}\right) \times \frac{1}{2 \sqrt{x}}=\frac{-\sec ^{2} \sqrt{x}}{4 \sqrt{x} \sqrt{\cot \sqrt{x}}}$

Differentiation of $y=\sqrt{\cot \sqrt{x}}$ is $\frac{-\sec ^{2} \sqrt{x}}{4 \sqrt{x} \sqrt{\cot \sqrt{x}}}$

 

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