If y = \frac{(x + 1)^3}{x^2 + 1}, find the value of \frac{dy}{dx} at x = 1. — JAMB UTME

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

Answer: 2

The numerical value of the derivative at

x = 1

is 2.

To find

\frac{dy}{dx}

x = 1

for

y = \frac{(x + 1)^3}{x^2 + 1}

, we use the quotient rule

\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

. Setting

u = (x + 1)^3

gives

\frac{du}{dx} = 3(x + 1)^2

, and setting

v = x^2 + 1

gives

\frac{dv}{dx} = 2x

. At

x = 1

u = 8

\frac{du}{dx} = 12

v = 2

, and

\frac{dv}{dx} = 2

. Substituting into the quotient formula gives

\frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2

Identify the components of the quotient rule

Let

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

and

v(x) = x^2 + 1

The given function is a quotient of two functions of

x

Differentiate the numerator using the chain rule and the denominator using standard rules

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

and

v'(x) = 2x

The chain rule states that

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

Evaluate all function components at

x = 1

u(1) = 8

u'(1) = 12

v(1) = 2

, and

v'(1) = 2

Substituting

x = 1

simplifies the calculation before applying the full quotient expression.

Apply the quotient rule formula to calculate the final derivative value

\frac{dy}{dx}\Big|_{x=1} = \frac{v(1)u'(1) - u(1)v'(1)}{[v(1)]^2} = \frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2

Substituting the numerical values yields the final result.

Combining the Quotient Rule and Chain Rule for differentiation

Question