Question

Q1.Given y =(x-4)/2sqrtx then dy/dx at x =1 =?

Answer 1

Step-by-step explanation:

Solution :

$\begin{gathered}\sf{Function\:of\:y = \dfrac{(x - 4)}{2\sqrt{x}}} \\ \\ \end{gathered} </p><p>$

By using the quotient rule of differentiation, we get :

$\begin{gathered}\underline{\sf{\bigstar\: Quotient\:rule\:of\: differentiation\: :-}} \\ \\ :\implies \sf{\dfrac{d}{dx}\left[\dfrac{v}{u}\right] = \dfrac{v\cdot \dfrac{du}{dx} - u\cdot \dfrac{dv}{dx}}{v^{2}}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d}{dx}\left[\dfrac{(x - 4)}{2\sqrt{x}}\right] = \dfrac{2\sqrt{x}\cdot \dfrac{d(x - 4)}{dx} - (x - 4)\cdot \dfrac{d(2\sqrt{x})}{dx}}{(2\sqrt{x})^{2}}} \\ \\ \end{gathered}$

By differentiating (x - 4) and (2√x), we get :

Differentiation of (x - 4) :

$\begin{gathered}\sf{Function\:of\:y = (x - 4)} \\ \\ \end{gathered} </p><p></p><p>$

By using the difference rule of differentiation, we get :

$\begin{gathered}\underline{\sf{\bigstar\: Difference\:rule\:of\: differentiation\: :-}} \\ \\ :\implies \sf{\dfrac{d}{dx}\left[f(x) - g(x)]\right] = \dfrac{d}{dx}[f(x)] - \dfrac{d}{dx}[g(x)]} \\ \\ \end{gathered} </p><p></p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d(x - 4)}{dx} = \dfrac{d(x)}{dx} - \dfrac{d(4)}{dx}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d(x - 4)}{dx} = 1 - 0} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}\underline{\over{\therefore \sf{\dfrac{d(x - 4)}{dx} = 1}}} \\ \\ \end{gathered} </p><p>$

Differentiation of 2√x :

$\begin{gathered}\sf{Function\:of\:y = 2\sqrt{x}\:or\:2\cdot x^{1/2}} \\ \\ \end{gathered} </p><p>$

By using the power rule of differentiation, we get :

$\begin{gathered}\underline{\sf{\bigstar\: Power\:rule\:of\: differentiation\: :-}} \\ \\ :\implies \sf{\dfrac{d(x^{n})}{dx} = n\cdot x^{(n - 1)}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d(2\sqrt{x})}{dx} = \dfrac{1}{2}\cdot 2 x^{(1/2 - 1)}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d(2\sqrt{x})}{dx} = x^{-1/2}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d(2\sqrt{x})}{dx} = \dfrac{1}{\sqrt{x}}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}\underline{\over{\therefore \sf{\dfrac{d(2\sqrt{x})}{dx} = \dfrac{1}{\sqrt{x}}}}} \\ \\ \end{gathered} </p><p>$

Now, by substituting the derivative of 2√x and (x - 4) in the equation, we get :

$\begin{gathered}:\implies \sf{\dfrac{d}{dx}\left[\dfrac{(x - 4)}{2\sqrt{x}}\right] = \dfrac{2\sqrt{x}\cdot \dfrac{d(x - 4)}{dx} - (x - 4)\cdot \dfrac{d(2\sqrt{x})}{dx}}{(2\sqrt{x})^{2}}} \\ \\ \end{gathered} </p><p></p><p> </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d}{dx}\left[\dfrac{(x - 4)}{2\sqrt{x}}\right] = \dfrac{2\sqrt{x}\cdot 1 - (x - 4)\cdot \dfrac{1}{\sqrt{x}}}{(2\sqrt{x})^{2}}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d}{dx}\left[\dfrac{(x - 4)}{2\sqrt{x}}\right] = \dfrac{2\sqrt{x} - \dfrac{(x - 4)}{\sqrt{x}}}{4x}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d}{dx}\left[\dfrac{(x - 4)}{2\sqrt{x}}\right] = \dfrac{\dfrac{2x - (x - 4)}{\sqrt{x}}}{4x}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{d}{dx}\left[\dfrac{(x - 4)}{2\sqrt{x}}\right] = \dfrac{x + 4}{4x\sqrt{x}}} \\ \\ \end{gathered} </p><p>$

Now, to find the derivative at x = 1 :

$\begin{gathered}:\implies \sf{\bigg(\dfrac{dy}{dx}\bigg)_{x = 1} = \dfrac{x + 4}{4x\sqrt{x}}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\bigg(\dfrac{dy}{dx}\bigg)_{x = 1} = \dfrac{1 + 4}{4(1)\sqrt{1}}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\bigg(\dfrac{dy}{dx}\bigg)_{x = 1} = \dfrac{5}{4}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}\boxed{\therefore \sf{\bigg(\dfrac{dy}{dx}\bigg)_{x = 1} = \dfrac{5}{4}}} \\ \\ \end{gathered} </p><p>$