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

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

112

Applying the product rule

\frac{dy}{dx} = u'v + uv'

along with the chain rule gives

u'(1) = 6

v(1) = 16

u(1) = 1

, and

v'(1) = 16

. Evaluating

u'(1)v(1) + u(1)v'(1)

yields

6 \times 16 + 1 \times 16 = 96 + 16 = 112

Decompose the function into two components for the Product Rule

Let

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

and

v(x) = (x^2 + 3)^2

, so

y = u(x) v(x)

The function is a product of two composite expressions.

Differentiate each term using the Chain Rule

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

v'(x) = 2(x^2 + 3) \cdot \frac{d}{dx}(x^2 + 3) = 4x(x^2 + 3)

Applying the Chain Rule requires differentiating the outer function and multiplying by the derivative of the inner function.

Apply the Product Rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 6(2x - 1)^2 (x^2 + 3)^2 + 4x(2x - 1)^3 (x^2 + 3)

The derivative of a product

u \cdot v

u'v + uv'

Evaluate the derivative at

x = 1

u(1) = (2(1) - 1)^3 = 1

u'(1) = 6(2(1) - 1)^2 = 6

v(1) = (1^2 + 3)^2 = 16

v'(1) = 4(1)(1^2 + 3) = 16

\left.\frac{dy}{dx}\right|_{x=1} = (6)(16) + (1)(16) = 96 + 16 = 112

Substitute

x = 1

into all expressions to find the numerical derivative.

Combined Product Rule and Chain Rule of Differentiation

Tahmini Süre:2m 0s

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