Question

Difficulty: MediumRules of Differentiation (Product, Quotient, and Chain Rules)

Given the function y=2x3(x2+1)2y = \frac{2x - 3}{(x^2 + 1)^2}, find the numerical value of dydx\frac{dy}{dx} at x=1x = 1.

Answer: 1

Answer

The numerical value of dydx\frac{dy}{dx} at x=1x = 1 is 1.
Applying the quotient rule dydx=uvuvv2\frac{dy}{dx} = \frac{u'v - uv'}{v^2} alongside the chain rule for the denominator yields dydx=2(x2+1)2(2x3)4x(x2+1)(x2+1)4\frac{dy}{dx} = \frac{2(x^2 + 1)^2 - (2x - 3) \cdot 4x(x^2 + 1)}{(x^2 + 1)^4}. Substituting x=1x = 1 evaluates to 2(4)(1)(8)16=1616=1\frac{2(4) - (-1)(8)}{16} = \frac{16}{16} = 1.

Step-by-Step Solution

1
Identify the numerator and denominator functions
Let u(x)=2x3u(x) = 2x - 3 and v(x)=(x2+1)2v(x) = (x^2 + 1)^2.
The given function is structured as a quotient y=uvy = \frac{u}{v}, requiring the quotient rule.
2
Find the derivatives u(x)u'(x) and v(x)v'(x)
u(x)=2u'(x) = 2 and v(x)=2(x2+1)2x=4x(x2+1)v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1).
Differentiating u(x)u(x) follows standard polynomial rules; v(x)v(x) requires the chain rule.
3
Evaluate u(1),u(1),v(1),u(1), u'(1), v(1), and v(1)v'(1) at x=1x = 1
u(1)=1u(1) = -1, u(1)=2u'(1) = 2, v(1)=4v(1) = 4, and v(1)=8v'(1) = 8.
Evaluating components before substitution simplifies the arithmetic.
4
Apply the quotient rule formula dydx=uvuvv2\frac{dy}{dx} = \frac{u'v - uv'}{v^2} at x=1x = 1
\frac{dy}{dx} = \frac{(2)(4) - (-1)(8)}{4^2} = \frac{8 + 8}{16} = 1.
Substitute the calculated component values into the quotient rule formula.

Key Concept

Quotient Rule and Chain Rule of Differentiation
Estimated Time:1m 30s
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