Question

differentiate w.r.t. x.
${( log_{e}x) }^{2}$
​

Answer 1

Answer:

Let,

$y = {( log_{e}x) }^{2}$

•:

$\frac{dy}{dx} = \frac{d}{dx} { (log_{}x) }^{2}$

$= > 2logx \: \frac{d}{dx} (logx)$

$= > 2logx. \frac{1}{x} = \frac{2}{x} logx$

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Answer 2

Answer: $\frac{2\log_2x}{x}$

Explanation:

Given,

$y=(\log_ex)^2$

$\text{Let}\;\;\log_ex=t\\\;\\\text{Diffrentiating it with respect to x}\\\\\;\frac{d\log_ex}{dx}=\frac{dt}{dx}\\\;\\\frac{1}{x}=\frac{dt}{dx}\\\;\\\frac{dt}{dx}=\frac{1}{x}$

Now,

$y=t^2\\\;\\\text{Differentiating it with respect to x}\\\;\\\frac{dy}{dx}=2t.\frac{dt}{dx}\\\;\\\frac{dy}{dx}=2\times\log_ex\times\frac{1}{x}\\\;\\\frac{dy}{dx}=\frac{2\log_ex}{x}$

 $\textbf{Note}:-\\\;\\1)\;\;\frac{d\log_ex}{dx}=\frac{1}{x}\\\;\\2)\;\;\frac{dx^n}{dx}=nx^{n-1}\\\;\\3)\;\;\text{If}\;\;y=f_1(u)\;,u=f_2(v)\;\;\text{and}\;,v=f_3(x)\\\;\\\implies\frac{du}{dx}=\frac{du}{dv}.\frac{dv}{dx}\\\;\\\implies\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dv}.\frac{dv}{dx}$