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Miscellaneous Exercise · Q9

Q.Find the derivative of px2+qx+rax+b\dfrac{px^2 + qx + r}{ax + b}.

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Apply the Quotient Rule to differentiate a rational function: the derivative of uv\frac{u}{v} is v⋅u′−u⋅v′v2\frac{v \cdot u' - u \cdot v'}{v^2}. The result is apx2+2bpx+bq−ar(ax+b)2\dfrac{apx^2 + 2bpx + bq - ar}{(ax+b)^2}.

When you have one polynomial divided by another, the Quotient Rule is your tool. The idea is simple: if y=u(x)v(x)y = \frac{u(x)}{v(x)}, then the rate of change of yy depends on how fast the numerator is changing relative to the denominator, adjusted for how the denominator itself is changing. The formula captures this interplay:

ddx(uv)=v⋅dudx−u⋅dvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}

The denominator v2v^2 appears because we're measuring the rate per unit of the "base" (the denominator), and both terms in the numerator account for the coupled changes.

Let me work through this problem step by step.

1. Identify the numerator and denominator

Here u(x)=px2+qx+ru(x) = px^2 + qx + r and v(x)=ax+bv(x) = ax + b.

2. Differentiate the numerator

dudx=ddx(px2+qx+r)=2px+q\frac{du}{dx} = \frac{d}{dx}(px^2 + qx + r) = 2px + q

3. Differentiate the denominator

dvdx=ddx(ax+b)=a\frac{dv}{dx} = \frac{d}{dx}(ax + b) = a

4. Apply the Quotient Rule

Substitute into the formula:

ddx(px2+qx+rax+b)=(ax+b)(2px+q)−(px2+qx+r)(a)(ax+b)2\frac{d}{dx}\left(\frac{px^2 + qx + r}{ax + b}\right) = \frac{(ax + b)(2px + q) - (px^2 + qx + r)(a)}{(ax + b)^2}

5. Expand the numerator

First term: (ax+b)(2px+q)(ax + b)(2px + q)

=ax⋅2px+ax⋅q+b⋅2px+b⋅q=2apx2+aqx+2bpx+bq= ax \cdot 2px + ax \cdot q + b \cdot 2px + b \cdot q = 2apx^2 + aqx + 2bpx + bq

Second term: (px2+qx+r)⋅a(px^2 + qx + r) \cdot a

=apx2+aqx+ar= apx^2 + aqx + ar …

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