If y = \frac{(2x^2 + 1)^3}{4x - 1}, determine the numerical value of \frac{dy}{dx} at x = … — JAMB UTME

If $y = \frac{(2x^2 + 1)^3}{4x - 1}$ , determine the numerical value of $\frac{dy}{dx}$ at $x = 1$ .

Cevap: 24

The numerical value of the derivative

\frac{dy}{dx}

x = 1

is 24.

Evaluating the derivative requires combined application of the quotient rule and chain rule. Differentiating

u(x) = (2x^2 + 1)^3

via the chain rule yields

u'(x) = 12x(2x^2 + 1)^2

. At

x = 1

u(1) = 27

u'(1) = 108

v(1) = 3

, and

v'(1) = 4

. Substituting these into the quotient rule

\frac{u'v - uv'}{v^2}

produces

\frac{324 - 108}{9} = 24

Decompose the function for the Quotient Rule

Let

u(x) = (2x^2 + 1)^3

and

v(x) = 4x - 1

The given expression is a quotient of two differentiable functions.

Differentiate the numerator using the Chain Rule

u'(x) = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2

The chain rule states that

\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)

Differentiate the denominator and evaluate all components at

x = 1

v'(1) = 4

v(1) = 3

u(1) = 27

, and

u'(1) = 108

Evaluating each term individually reduces algebraic complexity prior to applying the quotient rule.

Apply the Quotient Rule formula and evaluate

\frac{dy}{dx}\Big|_{x=1} = \frac{(108)(3) - (27)(4)}{3^2} = \frac{324 - 108}{9} = 24

The quotient rule formula is

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

Combination of Quotient Rule and Chain Rule

Tahmini Süre:1m 30s

Soru