The derivative of ln-1-x2-x-1-x2-x with respect to cos ln x is

Let y=ln(√(1+x^2)+x√(1+x^2) x) ⇒ y=ln(√(1+x^2)+x ) ln(√(1+x^2) x ) ⇒ nbsp;dydx=(1+2x2√(1+x^2))√(1+x^2)+x (2x2√(1+x^2) 1)√(1+x^2) x ⇒ nbsp;dydx=1√(1+x^2)+1√( ...

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Standard VI

Mathematics

Idea of a Set

The derivativ...

Question

The derivative of $ln (\frac{\sqrt{1 + x^{2}} + x}{\sqrt{1 + x^{2}} - x})$ with respect to $cos (ln x)$ is

\frac{2 x}{\sqrt{(1 + x^{2})} sin (ln x)}

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- \frac{2 x}{\sqrt{(1 + x^{2})} sin (ln x)}

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1

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- \frac{4 x}{\sqrt{(1 + x^{2})} sin (ln x)}

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Solution

The correct option is B $- \frac{2 x}{\sqrt{(1 + x^{2})} sin (ln x)}$
Let $y = ln (\frac{\sqrt{1 + x^{2}} + x}{\sqrt{1 + x^{2}} - x})$
$\Rightarrow y = ln (\sqrt{1 + x^{2}} + x) - ln (\sqrt{1 + x^{2}} - x) \Rightarrow \frac{d y}{d x} = \frac{(1 + \frac{2 x}{2 \sqrt{1 + x^{2}}})}{\sqrt{1 + x^{2}} + x} - \frac{(\frac{2 x}{2 \sqrt{1 + x^{2}}} - 1)}{\sqrt{1 + x^{2}} - x} \Rightarrow \frac{d y}{d x} = \frac{1}{\sqrt{1 + x^{2}}} + \frac{1}{\sqrt{1 + x^{2}}} \Rightarrow \frac{d y}{d x} = \frac{2}{\sqrt{1 + x^{2}}}$

Let $z = cos (ln x)$
$\Rightarrow \frac{d z}{d x} = - \frac{sin (ln x)}{x}$
Therefore, $\frac{d y}{d z} = \frac{d y}{d x} \times \frac{d x}{d z}$
$\Rightarrow \frac{d y}{d z} = - \frac{2 x}{\sqrt{1 + x^{2}} sin (ln x)}$

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Standard VI Mathematics

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The derivative of ln(√1+x2+x√1+x2−x) with respect to cos(lnx) is

The derivative of $ln (\frac{\sqrt{1 + x^{2}} + x}{\sqrt{1 + x^{2}} - x})$ with respect to $cos (ln x)$ is