Given the function y = x^2 \sqrt{3x^2 + 4}, find the numerical value of \frac{dy}{dx} at x… — JAMB UTME

Given the function $y = x^2 \sqrt{3x^2 + 4}$ , find the numerical value of $\frac{dy}{dx}$ at $x = 2$ .

Answer: 22

The numerical value of the derivative at

x = 2

is 22.

Applying both the Product Rule and Chain Rule correctly yields

\frac{dy}{dx} = 2x\sqrt{3x^2 + 4} + \frac{3x^3}{\sqrt{3x^2 + 4}}

. Evaluating this expression at

x = 2

gives

16 + 6 = 22

Decompose the function into a product of two functions

Let

u(x) = x^2

and

v(x) = \sqrt{3x^2 + 4} = (3x^2 + 4)^{1/2}

The function

y

is expressed as the product of

u(x)

and

v(x)

, requiring the Product Rule for differentiation.

Find the derivative of the inner square root function using the Chain Rule

v'(x) = \frac{1}{2}(3x^2 + 4)^{-1/2} \cdot 6x = \frac{3x}{\sqrt{3x^2 + 4}}

The Chain Rule states that

\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Combine derivatives using the Product Rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 2x\sqrt{3x^2 + 4} + \frac{3x^3}{\sqrt{3x^2 + 4}}

The Product Rule formula is

\frac{d}{dx}[u \cdot v] = u'v + uv'

Evaluate the derivative at

x = 2

\frac{dy}{dx}\Big|_{x=2} = 2(2)\sqrt{3(2)^2 + 4} + \frac{3(2)^3}{\sqrt{3(2)^2 + 4}} = 4(4) + \frac{24}{4} = 16 + 6 = 22

Substituting

x = 2

gives the specific slope of the tangent line to the curve at that point.

Combining the Product Rule and Chain Rule to evaluate derivatives of composite product functions

Question