Q.Differentiate $y = x^3+5x^2-7x+2$ with respect to $x$.

By the quotient rule, $\dfrac{d}{dx}\dfrac{1-3x}{1+3x}=\dfrac{(-3)(1+3x)-(1-3x)(3)}{(1+3x)^{2}}=\dfrac{-6}{(1+3x)^{2}}$.

This is a direct application of the quotient rule from the Differential Calculus chapter of the TN HSC Class-11 Business Mathematics syllabus.

Step 1 — Identify $u$ and $v$.

$u=1-3x\Rightarrow u'=-3;\qquad v=1+3x\Rightarrow v'=3.$

Step 2 — Quotient rule.

$\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{u'v-uv'}{v^{2}}.$

Step 3 — Substitute. …

Quotient rule: $\dfrac{u'v-uv'}{v^2}$ with $u=x^2+x+1,\ v=x^2-x+1$ gives $\dfrac{2(1-x^2)}{(x^2-x+1)^2}$.

A quotient-rule differentiation from the Differential Calculus unit of the Tamil Nadu HSC Class-11 Business Mathematics syllabus.

Step 1 — Set up. Let $u=x^2+x+1$ and $v=x^2-x+1$, so $u'=2x+1$ and $v'=2x-1$. The quotient rule gives

$\frac{dy}{dx}=\frac{u'v-uv'}{v^2}.$

Step 2 — Expand $u'v$.

$(2x+1)(x^2-x+1)=2x^3-x^2+x+1.$

Step 3 — Expand $uv'$.

$(x^2+x+1)(2x-1)=2x^3+x^2+x-1.$

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