Question

46. If y = log (log ,x), then
dy/dx
at x = e is
(a)1/e (b)2/e
(c) e (d) 0.​

Answer 1

Answer:

a)1/e

Step-by-step explanation:

a ) 1/e

plz send tnx

plz

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given that
• y = log ( log x )

To Find:

• We have to find the value of dy/dx

Concept Used:

Chain Rule of Differentiation :

If f and g both are differentiable function and F be a composite function defined by F ( x ) = f [ g ( x ) ]

Then differentiation of F( x ) is given by

$\large\boxed{\sf{\green{\dfrac{d}{dx}F ( x ) = \dfrac{d}{dx} f (g(x) ) \times \dfrac{d}{dx} g ( x ) }} }$

$\sf{ }$

Solution:

We have been given that

$\implies \sf{ y = log ( log \: x )}$

Let us assume log x = z

$\implies \sf{ y = log ( z )}$

Differentiating with respect to x

$\implies \sf{ \dfrac{dy}{dx} = \dfrac{d}{dx}( log \: z )}$

$\implies \sf{ \dfrac{dy}{dx} = \dfrac{1}{z} \times \dfrac{dz}{dx} }$

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

$\implies \sf{ \dfrac{dy}{dx} = \dfrac{1}{log \: x} \times \dfrac{1}{x} }$

Putting the finding at x = e

$\implies \sf{ \dfrac{dy}{dx} = \dfrac{1}{log \: e} \times \dfrac{1}{e} }$

$\implies \boxed{\sf{ \dfrac{dy}{dx} = \dfrac{1}{e} }}$

Hence Option A is correct

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$\huge\underline{\sf{\red{A} \orange{n} \green{s} \pink{w} \blue{e} \purple{r}}}$

$\large\boxed{\sf{\purple{\dfrac{dy}{dx} = \dfrac{1} {e} }} }$

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