You don't need implicit differentiation, part I

Calculus Level 2

If $y = \displaystyle \sqrt{\dfrac{x}{\sqrt{\dfrac{x}{\sqrt{\dfrac{x}{\sqrt{}...}}}}}}$ , find $\left. \dfrac{dy}{dx} \right|_{x= 8}$

-\dfrac{1}{12}

The derivative does not exist

\dfrac{4}{3}

\dfrac{1}{12}

-\dfrac{4}{3}

2 solutions

Rishabh Jain
Mar 19, 2016

$\Large y=x^{\overbrace{\small{\frac{1}{2}-\frac{1}{4}+\frac{1}{8}+\cdots}}^{\color{#D61F06}{\text{Infinite GP}}}}$

$\Large =x^{\small{\frac{\frac 12}{1+\frac 12}}}=x^{\frac 13}$

$\left. \dfrac{dy}{dx} \right|_{x= 8} =\left. \dfrac{1}{3}\left(\dfrac{1}{x^{\frac 23}} \right)\right|_{x= 8} =\boxed{\color{#EC7300}{\dfrac{1}{12}}}$

Cool one! I still used implicit differentiation :P

Sravanth C. - 5 years, 2 months ago

You're welcome to post what you did if you want! I could see where you did it but i didn't find it necessary.

Hobart Pao - 5 years, 2 months ago

Well, I used the most common method.

If $\large y=\sqrt { \frac { x }{ \sqrt { \frac { x }{ \sqrt { \frac { x }{ \sqrt { \frac { x }{ \sqrt { ... } } } } } } } } }$ $\Rightarrow y=\sqrt { \frac { x }{ y } } \Rightarrow { y }^{ 2 }=\frac { x }{ y } \Rightarrow { y }^{ 3 }=x\Rightarrow y=\sqrt [ 3 ]{ x }$ .

Now, $\large \frac { dy }{ dx } =\frac { d }{ dx } \sqrt [ 3 ]{ x } =\frac { 1 }{ 3\sqrt [ 3 ]{ { x }^{ 2 } } }$ . Plugging $8$ , we get $\frac{1}{12}$ .

Swapnil Das - 5 years, 2 months ago

@Swapnil Das – I did the same Swapnil. Cheers!

Sravanth C. - 5 years, 2 months ago

@Sravanth C. – Cheers! :)

Swapnil Das - 5 years, 2 months ago

@Swapnil Das – i dont find a single difference in my method. fully same . :))

Satyabrata Dash - 5 years, 2 months ago

@Satyabrata Dash – Oh, that's great :P

Swapnil Das - 5 years, 2 months ago

@Swapnil Das – Thats the standard method!

Nihar Mahajan - 5 years, 2 months ago

@Nihar Mahajan – Yep! The 'simple standard solution'.

Swapnil Das - 5 years, 2 months ago

Nice question.. It inspired me for this where I took this concept to the next level... Thanks for the inspiration anyways ... :-)

Rishabh Jain - 5 years, 2 months ago

Fahim Saikat
Apr 1, 2016

let, $y = \displaystyle \sqrt{\dfrac{x}{\sqrt{\dfrac{x}{\sqrt{\dfrac{x}{\sqrt{}...}}}}}}$ => $dfrac{x}{y} =y$ => $x=y^3$ => $y=x^\dfrac{1}{3}$

then, $\dfrac{dy}{dx} =\dfrac{1}{3} x^\dfrac{-2}{3}$

if x=8 then, $\dfrac{dy}{dx} =\dfrac{1}{3} x^\dfrac{-2}{3} = \dfrac{1}{12}$

You don't need implicit differentiation, part I

2 solutions

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