Given the function y = \frac{(x^2 + 1)^3}{2x - 3}, what is the numerical value of \frac{dy… — JAMB UTME

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

Answer: 50

The numerical value of

\frac{dy}{dx}

x = 2

is 50.

Applying the Quotient Rule

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

along with the Chain Rule to find

u'(x) = 6x(x^2 + 1)^2

, and evaluating all components at

x = 2

gives

u(2) = 125

u'(2) = 300

v(2) = 1

, and

v'(2) = 2

. Substituting these values yields

\frac{(300)(1) - (125)(2)}{1^2} = 50

Set up the Quotient Rule components

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

and

v(x) = 2x - 3

The given expression is a quotient of two functions requiring

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

Differentiate

u(x)

using the Chain Rule and

v(x)

using basic power rules

u'(x) = 6x(x^2 + 1)^2

and

v'(x) = 2

Differentiating the outer power 3 gives

3(x^2 + 1)^2

, and multiplying by the derivative of the inner function

(2x)

gives

6x(x^2 + 1)^2

Evaluate

u(2)

u'(2)

v(2)

, and

v'(2)

u(2) = 125

u'(2) = 300

v(2) = 1

, and

v'(2) = 2

Substituting

x = 2

into each function and derivative simplifies calculation of the overall derivative.

Substitute values into the Quotient Rule formula

\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(2)}{(1)^2} = 50

Evaluating

\frac{u'(2)v(2) - u(2)v'(2)}{[v(2)]^2}

yields the exact numerical result.

Combined Application of Quotient Rule and Chain Rule

Question