Question

solve :
$\sf{ \frac{d}{dx} \sqrt{ \sqrt{ \sqrt{ \sqrt{x} } } }}$
class 11 ..maths ...
derivatives​

Answer 1

Answer :

$\Rightarrow\mathsf{\dfrac{1}{16\:x^{\tiny{\dfrac{15}{16}}}}}$

Explanation :

Refer the attached image.

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Rules of Derivatives has been used. We are given a function. We see that we are give roots as the function of x. So first we can simplify the term of x and then use rules of derivative to find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{x^{1/2}\;=\;\bf{\sqrt{x}}}}$

$\\\;\boxed{\sf{(a^{m})^{n}\;=\;\bf{a^{mn}}}}$

$\\\;\boxed{\sf{x^{-1}\;=\;\bf{\dfrac{1}{x}}}}$

$\\\;\boxed{\sf{x^{n}\;=\;\bf{n\:x^{n\;-\;1}}}}$

This is the Power Rule of Derivatives.

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★ Solution :-

Given,

$\\\;\large{\sf{\mapsto\;\;\dfrac{d}{dx}\;\sqrt{\;\sqrt{\;\sqrt{\;\sqrt{\;x\;}}}}}}$

Firstly we need to simplify the value of x.

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~ For simplification of x ::

$\\\;\;\sf{\rightarrow\;\;\dfrac{d}{dx}\;\sqrt{\;\sqrt{\;\sqrt{\;\sqrt{\;x\;}}}}}$

Now using the First formula of exponents, we get,

$\\\;\;\bf{\Rightarrow\;\;\dfrac{d}{dx}\;\sqrt{\;\sqrt{\;\sqrt{\;(x^{1/2})\;}}}}$

$\\\;\;\bf{\Rightarrow\;\;\dfrac{d}{dx}\;\sqrt{\;\sqrt{\;(x^{1/2})^{1/2})\;}}}$

$\\\;\;\bf{\Rightarrow\;\;\dfrac{d}{dx}\;\sqrt{\;(x^{1/2})^{1/2})^{1/2}\;}}$

$\\\;\;\bf{\Rightarrow\;\;\dfrac{d}{dx}\;(x^{1/2})^{1/2})^{1/2})^{1/2}}$

Now after multiplying all these exponents, we get, and using the Second formula of exponents,

$\\\;\;\bf{\Rightarrow\;\;\dfrac{d}{dx}\;(x^{1/2})^{1/2})^{1/4}}$

$\\\;\;\bf{\Rightarrow\;\;\dfrac{d}{dx}\;(x^{1/2})^{1/8}}$

$\\\;\;\underline{\underline{\bf{\Rightarrow\;\;\dfrac{d}{dx}\;\bigg(x^{1/16}\bigg)}}}$

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~ For the Derivative of the Function ::

$\\\;\;\sf{\mapsto\;\;\dfrac{d}{dx}\;\sqrt{\;\sqrt{\;\sqrt{\;\sqrt{\;x\;}}}}}$

$\\\;\;\bf{\Longrightarrow\;\;\dfrac{d}{dx}\;\bigg(x^{1/16}\bigg)}$

Now using the Power Rule of Derivatives, we get,

$\\\;\;\sf{:\rightarrow\;\;x^{n}\;=\;\bf{n\:x^{n\;-\;1}}}$

$\\\;\;\sf{\Longrightarrow\;\;\dfrac{d}{dx}\;\bigg(x^{1/16}\bigg)\;\;=\;\bf{\dfrac{1}{16}\;x^{\bigg(\dfrac{1}{16}\;-\;1\bigg)}}}$

$\\\;\;\sf{\Longrightarrow\;\;\dfrac{d}{dx}\;\bigg(x^{1/16}\bigg)\;\;=\;\bf{\dfrac{1}{16}\;x^{\bigg(\dfrac{1\;-\;16}{16}\bigg)}}}$

$\\\;\;\sf{\Longrightarrow\;\;\dfrac{d}{dx}\;\bigg(x^{1/16}\bigg)\;\;=\;\bf{\dfrac{1}{16}\;x^{-15/16}}}$

Now using Third Formula of Exponents, we get,

$\\\;\;\sf{\Longrightarrow\;\;\dfrac{d}{dx}\;\bigg(x^{1/16}\bigg)\;\;=\;\bf{\dfrac{1}{16}\;\times\;\bigg(\dfrac{1}{x}\bigg)^{15/16}}}$

$\\\;\;\sf{\Longrightarrow\;\;\dfrac{d}{dx}\;\bigg(x^{1/16}\bigg)\;\;=\;\bf{\dfrac{1}{16\;\times\;x^{15/16}}}}$

Hence,

$\\\;\;\underline{\underline{\bf{\Longrightarrow\;\;\dfrac{d}{dx}\;\bigg(x^{1/16}\bigg)\;\;=\;\bf{\dfrac{1}{16\;\;x^{15/16}}\;\;or\;\;\dfrac{1}{16\;\;\sqrt[16]{x^{15}}}}}}}$

$\\\;\underline{\boxed{\tt{Thus,\;\;derivative\;\;of\;\;function\;=\;\bf{\dfrac{1}{16\;\;x^{15/16}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;fg\;=\;fg'\;+\;f'g}$

This is the Product Rule of derivatives.

$\\\;\sf{\leadsto\;\;f\;+\;g\;=\;f'\;+\;g'}$

This is the Sum Rule of derivatives.

$\\\;\sf{\leadsto\;\;f\;-\;g\;=\;f'\;-\;g'}$

This is the Difference Rule of derivatives.

$\\\;\sf{\leadsto\;\;\dfrac{f}{g}\;=\;\dfrac{(f'g\;-\;g'h)}{g^{2}}}$

This is the Quotient Rule of derivatives.

$\\\;\;\sf{\leadsto\;\;\dfrac{1}{f}\;=\;\dfrac{-f'}{f^{2}}}$

This is the Reciprocal Rule of derivatives.

$\\\;\;\sf{\leadsto\;\;\dfrac{dy}{dx}\;=\;\dfrac{dy}{du}\;\dfrac{du}{dx}}$

This is the Chain Rule of derivatives.