Question

find dy/DX to the function y = 2x-3/3x+4 at x=1​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{y = \frac{2x - 3}{3x + 4} }$

TO DETERMINE

$\displaystyle \sf{\frac{dy}{dx} \: \: at \: \: x = 1}$

EVALUATION

Here it is given that

$\displaystyle \sf{y = \frac{2x - 3}{3x + 4} }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = \frac{(3x + 4) \frac{d}{dx} (2x - 3) - (2x - 3) \frac{d}{dx} (3x + 4)}{ {(3x + 4)}^{2} } }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{2(3x + 4) -3 (2x - 3) }{ {(3x + 4)}^{2} } }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{6x + 8 -6x + 9 }{ {(3x + 4)}^{2} } }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{1 }{ {(3x + 4)}^{2} } }$

Putting x = 1 in both sides we get

$\displaystyle \sf{ \frac{dy}{dx} \bigg| _{x = 1} = \frac{1 }{ {(3.1 + 4)}^{2} } }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} \bigg| _{x = 1} = \frac{1 }{ {(7)}^{2} } }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} \bigg| _{x = 1} = \frac{1}{49} }$

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