Question

differentiate ( x+2) /(x+5) wrt x​

Answer 1

Solution:

$\implies \:\sf{y = \dfrac{x+2}{x+5}}$

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Remember:

Quotient rule of differentiation,

$\blacksquare \:\sf{\dfrac{dy}{dx}= \dfrac{v.\dfrac{du}{dx}- u.\dfrac{dv}{dx}}{v^2}}$

$\blacksquare$ Differentiation of x wrt x is 1.

$\blacksquare$Differntiation of constant wrt x is 0

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Differentiating wrt x,

$\implies\sf{\dfrac{dy}{dx} =\dfrac{(x+5). \dfrac{d(x+2)}{dx} - (x +2 )\dfrac{d(x+5)}{dx}}{{(x+5)}^2}}$

Differentiating (x+2) :

$\implies\sf{\dfrac{d(x+2)}{dx}= \dfrac{dx}{dx}+\dfrac{d2}{dx}}$

$\implies\sf{\dfrac{d(x+2)}{dx}= 1+0}$

$\implies\sf{\dfrac{d(x+2)}{dx}= 1}$

Differentiating (x+5):

$\implies\sf{\dfrac{d(x+5)}{dx}= \dfrac{dx}{dx}+\dfrac{d5}{dx}}$

$\implies\sf{\dfrac{d(x+5)}{dx}= 1+0}$

$\implies\sf{\dfrac{d(x+5)}{dx}= 1}$

Now,

$\implies\sf{\dfrac{dy}{dx} =\dfrac{(x+5) - (x +2 )}{{(x+5)}^2}}$

$\implies\sf{\dfrac{dy}{dx} =\dfrac{x+5 - x - 2 }{{(x+5)}^2}}$

$\implies\boxed{\sf{\dfrac{dy}{dx} =\dfrac{3 }{{(x+5)}^2}}}$

Answer 2

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

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$\longrightarrow \:(1). \: \boxed{ \red{\tt \: \dfrac{d}{dx}x = 1 }}$

$\longrightarrow \:(2). \: \boxed{ \red{\tt \: \dfrac{d}{dx}k \: = 0 }}$

$\longrightarrow \:(3). \: \boxed{ \red{\tt \: \dfrac{d}{dx}\dfrac{u}{v} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx} }{ {v}^{2} } }}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\tt \:   \longrightarrow \:Let \: y \: = \dfrac{x + 2}{x + 5}$

☆ Differentiate w. r. t. x both sides, we get

$\tt \:   \longrightarrow \:\dfrac{d}{dx}y = \dfrac{d}{dx} \bigg( \dfrac{x + 2}{x + 5} \bigg)$

$\tt \:   \longrightarrow \:\dfrac{dy}{dx} = \dfrac{(x + 5)\dfrac{d}{dx}(x + 2) - (x + 2)\dfrac{d}{dx}(x + 5)}{ {(x + 5)}^{2} }$

$\tt \:   \longrightarrow \:\dfrac{dy}{dx} = \dfrac{(x + 5)(1 + 0) - (x + 2)(1 + 0)}{ {(x + 5)}^{2} }$

$\tt \:   \longrightarrow \:\dfrac{dy}{dx} = \dfrac{ \cancel{x} + 5 - \cancel{x} - 2}{ {(x + 5)}^{2} }$

$\tt \:   \longrightarrow \:\dfrac{dy}{dx} = \dfrac{3}{ {(x + 5)}^{2} }$

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