Given that y = \frac{3x - 1}{(x + 2)^2}, what is the value of \frac{dy}{dx} at x = 1? — JAMB UTME

Given that $y = \frac{3x - 1}{(x + 2)^2}$ , what is the value of $\frac{dy}{dx}$ at $x = 1$ ?

The value of

\frac{dy}{dx}

x = 1

\frac{5}{27}

Applying the quotient rule

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

with

u = 3x - 1

and

v = (x + 2)^2

yields

\frac{dy}{dx} = \frac{8 - 3x}{(x + 2)^3}

. Substituting

x = 1

results in

\frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{27}

Identify the numerator

u(x)

and denominator

v(x)

for the quotient rule.

Let

u = 3x - 1

and

v = (x + 2)^2

The given function

y = \frac{u}{v}

requires the quotient rule

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

Differentiate

u(x)

and

v(x)

with respect to

x

u' = 3

and, using the chain rule,

v' = 2(x + 2)(1) = 2(x + 2)

The derivative of the inner term

(x + 2)

1

, giving

v' = 2(x + 2)

Substitute

u, v, u', v'

into the quotient rule formula and simplify.

\frac{dy}{dx} = \frac{3(x + 2)^2 - (3x - 1) \cdot 2(x + 2)}{(x + 2)^4} = \frac{(x + 2)[3(x + 2) - 2(3x - 1)]}{(x + 2)^4} = \frac{3x + 6 - 6x + 2}{(x + 2)^3} = \frac{8 - 3x}{(x + 2)^3}

Factoring out

(x + 2)

simplifies the algebraic expression.

Evaluate the derivative at

x = 1

\frac{dy}{dx}\Big|_{x=1} = \frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{3^3} = \frac{5}{27}

Substituting

x = 1

gives the final numerical derivative value.

Quotient and Chain Rules of Differentiation

Soru