If y = x^2 \sqrt{3x + 1}, what is the value of \frac{dy}{dx} evaluated at x = 1? — JAMB UTME

If $y = x^2 \sqrt{3x + 1}$ , what is the value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

The numerical value of

\frac{dy}{dx}

x = 1

\frac{19}{4}

Applying the product rule to

y = x^2 \sqrt{3x + 1}

gives

\frac{dy}{dx} = x^2 \cdot \frac{d}{dx}(\sqrt{3x + 1}) + \sqrt{3x + 1} \cdot \frac{d}{dx}(x^2)

. By the chain rule,

\frac{d}{dx}(\sqrt{3x + 1}) = \frac{3}{2\sqrt{3x + 1}}

. Substituting

x = 1

into

\frac{dy}{dx} = \frac{3x^2}{2\sqrt{3x + 1}} + 2x\sqrt{3x + 1}

gives

\frac{3}{4} + 4 = \frac{19}{4}

. Thus, the option equal to

\frac{19}{4}

is correct.

Identify the main differentiation rule required and split the function into two components.

Let

u = x^2

and

v = (3x + 1)^{1/2}

, so

y = u \cdot v

The function is a product of

x^2

and a composite function

\sqrt{3x + 1}

Differentiate

u

with respect to

x

, and use the chain rule to differentiate

v

with respect to

x

\frac{du}{dx} = 2x

and

\frac{dv}{dx} = \frac{1}{2}(3x + 1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x + 1}}

The chain rule requires multiplying the derivative of the outer square root function by the derivative of the inner linear function

3x + 1

Apply the product rule formula

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

\frac{dy}{dx} = x^2 \left( \frac{3}{2\sqrt{3x + 1}} \right) + \sqrt{3x + 1} (2x)

The derivative of a product

uv

u v' + v u'

Substitute

x = 1

into the derivative expression.

\frac{dy}{dx}\Big|_{x=1} = (1)^2 \left( \frac{3}{2\sqrt{3(1) + 1}} \right) + \sqrt{3(1) + 1} \cdot 2(1) = 1 \cdot \left( \frac{3}{4} \right) + 2 \cdot 2 = \frac{3}{4} + 4 = \frac{19}{4}

Evaluating at

x = 1

requires simplifying

\sqrt{4} = 2

and combining fractions.

Combining Product Rule and Chain Rule for Differentiation

Question