Question

Prove that the derivative of Ln(x) = 1/x from the definition of implicit differentiation.​​

Answer 1

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Answer 2

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

Let assume that,

$\rm :\longmapsto\:y = ln(x)$

So,

$\rm :\longmapsto\: {e}^{y} = {e}^{ln(x)}$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {e}^{ln(x)} \: = \: x \: }}}$

So,

$\rm :\longmapsto\:{e}^{y} = x$

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

$\rm :\longmapsto\:\dfrac{d}{dx}{e}^{y} = \dfrac{d}{dx}x$

We know,

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

So, using these, we get

$\rm :\longmapsto\:{e}^{y}\dfrac{dy}{dx} = 1$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{{e}^{y}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{x} \: \: \: \: \: \: \{ \because \: {e}^{y} = x \}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \dfrac{d}{dx} \: ln(x) \: = \: \frac{1}{x} \: }}$

Hence, Proved

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More to know :-

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