Rules of Differentiation (Product, Quotient, and Chain Rules)

21 questions

Question 21Question

If y=(2x2+1)3y = (2x^2 + 1)^3, what is the numerical value of dydx\frac{dy}{dx} evaluated at x=1x = 1?

Show answer & explanation

Answer: 108108

Answer

108108
Differentiating y=(2x2+1)3y = (2x^2 + 1)^3 using the chain rule yields dydx=3(2x2+1)24x=12x(2x2+1)2\frac{dy}{dx} = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2. Substituting x=1x = 1 gives 12(1)(3)2=10812(1)(3)^2 = 108.

Step-by-Step Solution

1
Identify inner and outer functions for the composite expression
Let u=2x2+1u = 2x^2 + 1, so y=u3y = u^3.
The function requires the application of the Chain Rule.
2
Compute the derivatives of the outer and inner functions
dydu=3u2=3(2x2+1)2\frac{dy}{du} = 3u^2 = 3(2x^2 + 1)^2 and dudx=4x\frac{du}{dx} = 4x.
Apply the power rule to both components.
3
Apply the Chain Rule formula dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
dydx=3(2x2+1)2(4x)=12x(2x2+1)2\frac{dy}{dx} = 3(2x^2 + 1)^2 \cdot (4x) = 12x(2x^2 + 1)^2.
Multiply the outer derivative by the inner derivative.
4
Substitute x=1x = 1 into the derivative
dydxx=1=12(1)(2(1)2+1)2=12(3)2=129=108\frac{dy}{dx}\Big|_{x=1} = 12(1)(2(1)^2 + 1)^2 = 12(3)^2 = 12 \cdot 9 = 108.
Evaluate at the given x-value.

Key Concept

Chain Rule of Differentiation
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