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If Y = X Tan X + √ X 2 + 1 2 , Find D Y D X ? - Mathematics

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प्रश्न

\[\text{ If }y = x^{\tan x} + \sqrt{\frac{x^2 + 1}{2}}, \text{ find} \frac{dy}{dx}\] ?

 

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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अध्याय 11: Differentiation - Exercise 11.05 [पृष्ठ ९०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.05 | Q 60 | पृष्ठ ९०

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