Question

Differentiate the following function with respect to x

y = log(x + sqrt{x^2 + 1})

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y = log[ \: x + \sqrt{ {x}^{2} + 1 } \: ]$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx} y =\dfrac{d}{dx} log[ \: x + \sqrt{ {x}^{2} + 1 } \: ]$

We know,

$\boxed{\tt{ \dfrac{d}{dx}logx = \frac{1}{x} \: }}$

So, using this result, we get

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + 1} } \dfrac{d}{dx}(x + \sqrt{ {x}^{2} + 1})$

We know,

$\boxed{\tt{ \dfrac{d}{dx}x = 1}}$

and

$\boxed{\tt{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }}$

So, using these results, we get

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + 1} }\bigg(1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1}}\dfrac{d}{dx}( {x}^{2} + 1) \bigg)$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

So, using this result, we get

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + 1} }\bigg(1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1}}(2x + 0)\bigg)$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + 1} }\bigg(1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1}}(2x)\bigg)$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + 1} }\bigg(1 + \dfrac{x}{ \sqrt{ {x}^{2} + 1}}\bigg)$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + 1} }\bigg(\dfrac{ \sqrt{ {x}^{2} + 1 } + x}{ \sqrt{ {x}^{2} + 1}}\bigg)$

$\rm\implies \:\boxed{\tt{ \dfrac{dy}{dx} = \frac{1}{ \sqrt{ {x}^{2} + 1 } } \: }}$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$