CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x: $\text{e}^{\sqrt{\cot\text{x}}}$
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Answer
Let, $\text{y}=\text{e}^\sqrt{{\cot\text{x}}}$ $\Rightarrow\ \text{y}=\text{e}^{(\cot\text{x})^\frac{1}{2}}$ Differentiate with respect to x we get, $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{e}^{(\cot\text{x})^\frac{1}{2}}\Big)$ $=\text{e}^{(\cot\text{x})^\frac{1}{2}}\frac{\text{d}}{\text{dx}}(\cot\text{x})^\frac{1}{2}$ [Using chain rule] $=\text{e}^\sqrt{\cot\text{x}}\times\frac{1}{2}(\cot\text{x})^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}(\cot\text{x})$ $=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$ So, $\frac{\text{d}}{\text{dx}}\Big(\text{e}^\sqrt{\cot\text{x}}\Big)=-\frac{\text{e}^\sqrt{\cot\text{x}}\times\text{cosec}^2\text{x}}{2\sqrt{\cot\text{x}}}$
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