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Worked solution

Keep the quotient-rule numerator order

Differentiate 2x3+3x4+2\dfrac{2x^3+3}{x^4+2} with respect to xx.

CJC · O Level revision · Example 26

Answer

2x2(−x4−6x+6)(x4+2)2\dfrac{2x^2(-x^4-6x+6)}{(x^4+2)^2}.

Compare your reasoning

Apply denominator times numerator derivative minus the reverse

ddx(2x3+3x4+2)=(x4+2)ddx(2x3+3)−(2x3+3)ddx(x4+2)(x4+2)2\begin{aligned}&\frac{d}{dx}\left(\frac{2x^3+3}{x^4+2}\right)\\ & =\frac{(x^4+2)\frac d{dx}(2x^3+3)-(2x^3+3)\frac d{dx}(x^4+2)}{(x^4+2)^2}\end{aligned}

The denominator x4+2x^4+2 is positive for all real x. Keep the squared denominator and correct subtraction order.

Substitute and factor the numerator

(x4+2)(6x2)−(2x3+3)(4x3)(x4+2)2=2x2[(3x4+6)−(4x4+6x)](x4+2)2=2x2(−x4−6x+6)(x4+2)2.\begin{aligned}\frac{(x^4+2)(6x^2)-(2x^3+3)(4x^3)}{(x^4+2)^2}&=\frac{2x^2[(3x^4+6)-(4x^4+6x)]}{(x^4+2)^2}\\&=\frac{2x^2(-x^4-6x+6)}{(x^4+2)^2}.\end{aligned}

Preserve both source factorisation stages, including the 6x term after extracting 2x².

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