Question

Differentiate the following with respect to x : $\implies\mathtt{log_{e^{2} } logx}$

Answer 1

Answer is given in the pic.

Answer 2

We have to differentiate the following expression with respect to x :-

$\implies\mathtt{log_{e^{2} } logx}$

we can also write the above expression as :-

$\implies \dfrac{1}{2} \: \mathtt{log_{e} logx}$

Now :-

$\implies \dfrac{d(\dfrac{1}{2} \: \mathtt{log_{e} logx})}{dx}$

$\implies \dfrac{1}{2} \dfrac{d( \mathtt{log_{e} logx})}{dx}$

Now using Chain rule :-

$\implies \dfrac{1}{2} \dfrac{d( \mathtt{log_{e} logx})}{dx}$

$\implies \dfrac{1}{2} \dfrac{ \mathtt{1}}{logx} \times \dfrac{ \mathtt{d(logx)}}{dx} \times \dfrac{dx}{dx}$

$\implies \dfrac{1}{2} \dfrac{ \mathtt{1}}{logx} \times \dfrac{ \mathtt{1}}{x} \times \dfrac{1}{1}$

$\implies \dfrac{ \mathtt{1}}{2xlogx}$

Therefore :-

$\implies \dfrac{ \mathtt{dy}}{dx}= \dfrac{ \mathtt{1}}{2xlogx}$

Answer :

$\dfrac{ \mathtt{1}}{2xlogx}$

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d(sinx)/dx = cosx

d(cos x)/dx = -sin x

d(cosec x)/dx = -cot x cosec x

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x

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