Question

Find
$finde \: \frac{dy}{dx \: } \: if \: y = log( logx)$
​

Answer 1

Given :

$y = log( logx)$

To find :

$\dfrac{dy}{dx}$

Solution :

$y = log( logx)$

Now differentiating both sides :-

$\implies\dfrac{dy}{dx}= \dfrac{d(log( logx))}{dx}$

We know that :-

• d(log t)/dt = 1/t

Now using Chain rule :-

$\implies\dfrac{dy}{dx}= \dfrac{1}{ log(x) } \times \dfrac{d( log(x) )}{dx} \times \dfrac{dx}{dx}$

$\implies\dfrac{dy}{dx}= \dfrac{1}{ log(x) } \times \dfrac{1}{x} \times 1$

$\implies\dfrac{dy}{dx}= \dfrac{1}{x} \dfrac{1}{ log(x) }$

Answer :

$\dfrac{dy}{dx}= \dfrac{1}{x} \dfrac{1}{ log(x) }$

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$\large\bf\blue{Additional-Information}$

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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Answer 2

Given ,

The function is $\tt y = log \{log(x) \}$

Differentiaing y wrt x by using chain rule , we get

$\tt \frac{dy}{dx} = \frac{d}{dx} log \{ log(x) \}$

$\tt \frac{dy}{dx} = \frac{1}{ log(x) } \frac{d}{dx} log(x)$

$\tt \frac{dy}{dx} = \frac{1}{ log(x) } \times \frac{1}{x}$

$\tt \frac{dy}{dx} = \frac{1}{x log(x) }$

Remmember :

$\star \: \: \tt \frac{d}{dx} log(x) = \frac{1}{x}$

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