Given the function y = \frac{(x^2 + 1)^3}{3x - 5}, calculate the value of \frac{dy}{dx} at… — JAMB UTME

Given the function $y = \frac{(x^2 + 1)^3}{3x - 5}$ , calculate the value of $\frac{dy}{dx}$ at $x = 2$ .

Cevap: -75

The value of

\frac{dy}{dx}

x = 2

-75

Applying the Quotient Rule

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

along with the Chain Rule to differentiate

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

yields

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

. Evaluating at

x=2

gives

u(2)=125

u'(2)=300

v(2)=1

, and

v'(2)=3

, leading to

\frac{300(1) - 125(3)}{1^2} = -75

Set up the Quotient Rule framework

Let

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

and

v(x) = 3x - 5

, so that

y = \frac{u(x)}{v(x)}

and

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

The function is expressed as a quotient of two differentiable terms.

Differentiate the numerator using the Chain Rule

u'(x) = 3(x^2 + 1)^2 \cdot \frac{d}{dx}(x^2 + 1) = 6x(x^2 + 1)^2

The numerator is a composite function requiring the inner derivative derivative of

x^2+1

to be multiplied.

Differentiate the denominator

v'(x) = 3

The derivative of a linear function

3x - 5

with respect to

x

is its coefficient 3.

Evaluate

u(x)

u'(x)

v(x)

, and

v'(x)

x = 2

u(2) = 125

u'(2) = 300

v(2) = 1

v'(2) = 3

Substituting

x = 2

into each evaluated component simplifies the numerical calculation.

Substitute numerical values into the Quotient Rule formula

\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(3)}{1^2} = \frac{300 - 375}{1} = -75

Completing the arithmetic calculation yields the final numerical derivative value.

Combining the Quotient Rule and Chain Rule for composite fractional functions

Tahmini Süre:1m 30s

Soru