Question

Question :-

Find derivative of given functions w.r.t independent variable Y = 2x + 5/ 3x - 2

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Answer 1

$\underline\red{{\rm Answer - }}$

$\rm \dfrac{dy}{dx} = \dfrac{ - 19}{{(3x - 2)}^{2}}$

$\underline{\bf Solution - }$

$\underline{\sf Given - }$

Function of dependent variable with respect to independent variable -

$\rm Y = \dfrac{2x \: + \: 5}{3x \: - \: 2}$

• $\rm u = 2x + 5 ( let )$
• $\rm v = 3x - 2 ( let )$

$\underline{\bf To \:find - }$

Derivative of the function - $\sf \dfrac{dy}{dx}$

$\underline{\bf Formula \:used - }$

Quotient rule -

$\boxed{\sf \gray{ \dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{ v \Bigg(\dfrac{du}{dx}\Bigg) - u\dfrac{dv}{dx}}{{v}^{2}}}}$

$\boxed{\sf \gray{ \dfrac{d(c)}{dx} = 0}}$

where c is any constant.

━━━━━━━━━━━

$\sf\implies \dfrac{du}{dx} = \dfrac{d(2x + 5)}{dx}$

$\sf \implies\dfrac{d(2x^{1} + 5)}{dx}$

$\sf \implies 2 + 0$

$\implies \sf\frac{du}{dx} = 2$

$\sf \implies \dfrac{dv}{dx} = \dfrac{d(3x - 2)}{dx}$

$\sf \implies \dfrac{d(3x^{1} - 2)}{dx}$

$\sf \implies 3 - 0$

$\sf \implies \dfrac{dv}{dx} = 3$

• $\tt u = 2x + 5$
• $\tt v = 3x - 2$

• $\tt \dfrac{du}{dx} = 2$

• $\tt \dfrac{dv}{dx} = 3$

Substituting the value in formula -

$\sf \dfrac{d}{dx}(\frac{u}{v}) = \frac{ v (\frac{du}{dx}) - u\frac{dv}{dx}}{{v}^{2}}$

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

$\sf \implies \dfrac{6x - 4 - (6x + 15)}{{(3x - 2)}^{2}}$

$\sf \implies \dfrac{ \cancel{6x} - \: 4 \: \cancel{- 6x} - 15}{{(3x - 2)}^{2}}$

$\sf \implies \dfrac{ - 19}{{(3x - 2)}^{2}}$

$\sf \implies \dfrac{ - 19}{{(3x - 2)}^{2}}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{ - 19}{{(3x - 2)}^{2}}$