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Given y = (x² + 3)/(x³ + 4).

March 29, 2026Mathematics

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Answer

\frac{-2}{25}

Graph of y = (x² + 3)/(x³ + 4)

Key points of y = (x² + 3)/(x³ + 4)
Point	x	y
Minimum	0	0.75
Maximum	0.83	0.81

5 steps

Identify $u$ and $v$ for the quotient rule.
Given $y = \frac{x^2 + 3}{x^3 + 4}$ .
Let $u = x^2 + 3$ and $v = x^3 + 4$ .
Find the derivatives of $u$ and $v$ with respect to $x$ .
$\frac{du}{dx} = \frac{d}{dx}(x^2 + 3) = 2x$ $\frac{dv}{dx} = \frac{d}{dx}(x^3 + 4) = 3x^2$
Apply the quotient rule.
The quotient rule states that if $y = \frac{u}{v}$ , then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$ .
Substitute the expressions for $u, v, \frac{du}{dx}, \frac{dv}{dx}$ :
$\frac{dy}{dx} = \frac{(x^3 + 4)(2x) - (x^2 + 3)(3x^2)}{(x^3 + 4)^2}$
Simplify the expression for $\frac{dy}{dx}$ .
Expand the numerator:
$\frac{dy}{dx} = \frac{2x^4 + 8x - (3x^4 + 9x^2)}{(x^3 + 4)^2}$ $\frac{dy}{dx} = \frac{2x^4 + 8x - 3x^4 - 9x^2}{(x^3 + 4)^2}$
Combine like terms in the numerator:
$\frac{dy}{dx} = \frac{-x^4 - 9x^2 + 8x}{(x^3 + 4)^2}$
Evaluate $\frac{dy}{dx}$ at $x=1$ .
Substitute $x=1$ into the derivative expression:
$\frac{dy}{dx}\Big|_{x=1} = \frac{-(1)^4 - 9(1)^2 + 8(1)}{((1)^3 + 4)^2}$ $= \frac{-1 - 9 + 8}{(1 + 4)^2}$ $= \frac{-10 + 8}{(5)^2}$ $= \frac{-2}{25}$
The value of $\frac{dy}{dx}$ at $x=1$ is:
$\frac{-2}{25}$

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