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Zorluk: KolayRules of Differentiation (Product, Quotient, and Chain Rules)

If y=4x+1x−2y = \frac{4x + 1}{x - 2}, calculate the numerical value of dydx\frac{dy}{dx} at x=3x = 3.

Cevap: -9

Cevap

The numerical value of dydx\frac{dy}{dx} at x=3x = 3 is −9-9.
Using the quotient rule dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} for y=4x+1x−2y = \frac{4x + 1}{x - 2} gives dydx=(x−2)(4)−(4x+1)(1)(x−2)2=−9(x−2)2\frac{dy}{dx} = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} = \frac{-9}{(x - 2)^2}. Substituting x=3x = 3 produces −9(3−2)2=−9\frac{-9}{(3 - 2)^2} = -9.

Adım Adım Çözüm

1
Identify u(x)u(x) and v(x)v(x) for the quotient rule formula.
u=4x+1u = 4x + 1 and v=x−2v = x - 2.
The given equation y=4x+1x−2y = \frac{4x + 1}{x - 2} is a quotient of two functions of xx.
2
Calculate the individual derivatives dudx\frac{du}{dx} and dvdx\frac{dv}{dx}.
dudx=4\frac{du}{dx} = 4 and dvdx=1\frac{dv}{dx} = 1.
These derivatives are required components of the quotient rule.
3
Substitute the expressions into the quotient rule formula dydx=vu′−uv′v2\frac{dy}{dx} = \frac{v u' - u v'}{v^2} and simplify.
dydx=(x−2)(4)−(4x+1)(1)(x−2)2=4x−8−4x−1(x−2)2=−9(x−2)2\frac{dy}{dx} = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} = \frac{4x - 8 - 4x - 1}{(x - 2)^2} = \frac{-9}{(x - 2)^2}.
Simplifying the numerator yields the general derivative function.
4
Evaluate the derivative at x=3x = 3.
dydx∣x=3=−9(3−2)2=−91=−9\frac{dy}{dx}\Big|_{x=3} = \frac{-9}{(3 - 2)^2} = \frac{-9}{1} = -9.
Substituting x=3x = 3 gives the requested numerical value.

Anahtar Kavram

Quotient Rule of Differentiation
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