Question

differentiate the following​

Answer 1

Answer:

Given that,

• $\tt{y =\dfrac{x-4}{\sqrt{x+2}}}$

We need to find,

• $\tt{y =\dfrac{dy}{dx}}$

Solution:-

The Given value of y is in division, so we will follow the $\sf{\dfrac{du}{dv}}$ method.

By this method, $\sf{\dfrac{dy}{dx} = \dfrac{\dfrac{vdu}{dx}-\dfrac{udv}{dx}}{v^{2}}}$

So, let us take u = x-4 and v = √(x-2)

Let us first find:- du/dx and dv/dx.

As u = x-4.

Hence, $\sf{\dfrac{du}{dx}=1}$

Now, As $\sf{v=\sqrt{x+2}}$.

So, $\sf{\dfrac{dv}{dx}=\dfrac{1}{2\sqrt{x+2}}}$

So, substituting the values.

$\longrightarrow\tt{\dfrac{dy}{dx}=\dfrac{(\sqrt{x+2} \times 1) - (x-4 \times \dfrac{1}{2\sqrt{2+x}})}{{(\sqrt{x+2})}^{2}}}$

$\longrightarrow\tt{\dfrac{dy}{dx}=\dfrac{(\sqrt{x+2})-\dfrac{x-4}{2\sqrt{x+2}}}{x+2}}$

Separating,

$\longrightarrow\tt{\dfrac{dy}{dx}=\dfrac{\sqrt{x+2}}{x+2} -\dfrac{\dfrac{x-4}{2\sqrt{x+2}}}{x+2}}$

$\longrightarrow\tt{\dfrac{dy}{dx}=\dfrac{\sqrt{x+2}}{x+2}-\dfrac{x-4}{2\sqrt{x+2}\times (x+2)}}$

Simplifying further,

$\longrightarrow\tt{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x+2}}-\dfrac{x-4}{2\times {(x+2)}^{\frac{1}{2}}\times (x+2)}}$

$\longrightarrow\tt{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{x+2}}-\dfrac{x-4}{2\times {(x+2)}^{\frac{3}{2}}}}$

$\rule{200}2$

Answer 2

Answer:

Given:

We have been provided with a function such that :

$y = \dfrac{ x - 4}{ \sqrt{x + 2} }$

We need to find the derivative of y wrt x . Since the function is given in a fraction form, we need to apply Quotient Rule of Differentiation.

$\dfrac{dy}{dx} = \dfrac{ \bigg \{( \sqrt{x + 2} )\times \dfrac{d(x - 4)}{dx} \bigg \} - \bigg \{(x - 4) \times \dfrac{d( \sqrt{x + 2} )}{dx} \bigg \} }{ { (\sqrt{x + 2} )}^{2} }$

$= > \dfrac{dy}{dx} = \dfrac{ \bigg \{( \sqrt{x + 2}) \times 1 \bigg \} - \bigg \{(x - 4) \times \dfrac{1}{2 \sqrt{x + 2} } \bigg \} }{x + 2}$

Separating each numerator term :

$= > \dfrac{dy}{dx} = \dfrac{ \sqrt{x + 2} }{x + 2} - \dfrac{ \bigg( \dfrac{x - 4}{2 \sqrt{x + 2} } \bigg) }{x + 2}$

$= > \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{x + 2} } - \dfrac{x - 4}{2 {(x + 2)}^{ \frac{3}{2} } }$

So final answer is :

$\boxed{ \red{ \large{ \bold{ \dfrac{dy}{dx} = \dfrac{1}{ \sqrt{x + 2} } - \dfrac{x - 4}{2 {(x + 2)}^{ \frac{3}{2} } } }}}}$

Basic formulas used:

$1) \: \: y = {x}^{n}$

$= > \dfrac{dy}{dx} = n {x}^{n - 1}$

$2) \: \: y = \dfrac{u}{v}$

$= > \dfrac{dy}{dx} = \dfrac{v \frac{du}{dx} - u\frac{dv}{dx} }{ {v}^{2} }$