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प्रश्न
उत्तर
\[y = x^{\tan x } + \sqrt{\frac{x^2 + 1}{2}}\]
\[\text{ Taking log on both sides, we get }\]
\[\log y = \tan x\log x + \frac{1}{2}\log\left( \frac{x^2 + 1}{2} \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{\tan x}{x} + \sec^2 x\log x + \frac{1}{2} \times \frac{2}{x^2 + 1} \times x\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{\tan x}{x} + \sec^2 x\log x + \frac{x}{x^2 + 1}\]
\[ \Rightarrow \frac{dy}{dx} = \left( x^{\tan x } + \sqrt{\frac{x^2 + 1}{2}} \right)\left( \frac{\tan x}{x} + \sec^2 x\log x + \frac{x}{x^2 + 1} \right)\]
\[ \Rightarrow \frac{dy}{dx} = x^{\tan x } \left( \frac{\tan x}{x} + \sec^2 x\log x \right) + \frac{x}{\sqrt{2 x^2 + 2}}\]
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