Question

$\large{\red{\bf Question }}$

if $\sf y = x^\frac{1}{2}$

find the value of $\sf\dfrac{dy}{dx}$

Class - 11th

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Answer 1

$y = {x}^{ \frac{1}{2} }$

By Using,

$\dfrac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1}$

We get,

$\dfrac{dy}{dx} = \dfrac{1}{2} {x}^{ \frac{1}{2} - 1}$

$\dfrac{dy}{dx} = \dfrac{1}{2} {x}^{ \frac{ - 1}{2} }$

$\boxed{ \dfrac{dy}{dx} = \dfrac{1}{ 2\sqrt{x} } }$

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Irony: 9th moved to 10th Grader.

Answer 2

$\\ \underbrace{ \underline \color{grey}\large \sf \: Solution :- }$

It is also possible to find this derivative using the definition of the derivative, or

$\\ \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: f' (x) \: = \: \lim_{h \rightarrow0 } \: \dfrac{f(x + h) - f(x)}{h}$

It's worth noting that $\sf \:y = x^{ \frac{1}{2} } = \sqrt{x} :$

Now,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \dfrac{dy}{dx} = \dfrac{d}{dx} {x}^{ \frac{1}{2} }$

$\\ \: \: \therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: = \frac{d}{dx} \sqrt{x}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: = \: \lim_{h \rightarrow0 } \: \dfrac{ \sqrt{ (x + h)} - \sqrt x}{h}$

At first, it may appear as though there is no obvious simplification, but one trick we can use is to Rationalize the denominator.

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: = \: \lim_{h \rightarrow0 } \: \dfrac{ (\sqrt{ (x + h)} - \sqrt{x}) \: \: ( \sqrt{(x + h)} + \sqrt{x} ) } {h( \sqrt{(x + h)} + \sqrt{x} ) }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: = \: \lim_{h \rightarrow0 } \: \dfrac{ x + h + \sqrt{x (x + h)} - \sqrt{x(x + h)} } {h( \sqrt{x + h} + \sqrt{x} ) }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: = \: \lim_{h \rightarrow0 } \: \dfrac{ h } {h( \sqrt{(x + h)} + \sqrt{x} ) }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: = \: \lim_{h \rightarrow0 } \: \: \: \: \dfrac{ 1 } {( \sqrt{(x + h)} + \sqrt{x} ) }$

Now , here

you will notice that , when h = 0 is substituted into the limit, we no longer have an indeterminate form, meaning we can now find derivative by simple substitution.

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{\boxed{\: \bold{\: \: \dfrac{dy}{dx} = \dfrac{1}{ 2\sqrt{ x}} \: \: }} }$

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