For y=3x+5x−3, find the instantaneous rate of change when x=2

Step 1 y=3x+5x−3 dydx=(x−3)ddx(3x+5)−(3x+5)ddx(x−3)(x−3)2 Step 2 dydx=(x−3)3−(3x+5)(x−3)2 dydx=3x−9−3x−5(x−3)2 dydx=−14(x−3)2 (dydx)x=2=−14(2−3)2 (dydx)x=2=−14 So instantaneous rate of change =-14

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Ernstfalld

Answered question

2021-09-19

For

y = \frac{3 x + 5}{x - 3}

, find the instantaneous rate of change when x=2

Answer & Explanation

Anonym

Skilled2021-09-20Added 108 answers

Step 1

y = \frac{3 x + 5}{x - 3}

\frac{d y}{d x} = \frac{(x - 3) \frac{d}{d x} (3 x + 5) - (3 x + 5) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}

Step 2

\frac{d y}{d x} = \frac{(x - 3) 3 - (3 x + 5)}{{(x - 3)}^{2}}

\frac{d y}{d x} = \frac{3 x - 9 - 3 x - 5}{{(x - 3)}^{2}}

\frac{d y}{d x} = \frac{- 14}{{(x - 3)}^{2}}

{(\frac{d y}{d x})}_{x = 2} = \frac{- 14}{{(2 - 3)}^{2}}

{(\frac{d y}{d x})}_{x = 2} = - 14

So instantaneous rate of change =-14

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