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Miscellaneous Exercise · Q3

Q.Find the derivative of (px+q)(rx+s)(px + q)\left(\dfrac{r}{x} + s\right).

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✓ Free question

Expand the product first, then differentiate term by term using the power rule. The derivative is ps−prx2ps - \dfrac{pr}{x^2}.

When faced with a product of two functions, you have two roads: apply the Product Rule directly, or expand and differentiate. Here, expansion is cleaner because the algebra is straightforward and reveals the structure immediately.

The expression (px+q)(rx+s)(px + q)\left(\dfrac{r}{x} + s\right) is a product, but each factor is simple. Multiplying out transforms it into a sum of power functions, each of which differentiates in one step.

Step-by-step solution

  1. Expand the product Distribute (px+q)(px + q) across (rx+s)\left(\dfrac{r}{x} + s\right):

(px+q)(rx+s)=px⋅rx+px⋅s+q⋅rx+q⋅s(px + q)\left(\dfrac{r}{x} + s\right) = px \cdot \dfrac{r}{x} + px \cdot s + q \cdot \dfrac{r}{x} + q \cdot s

Simplify each term:

=pr+psx+qrx+qs= pr + psx + \dfrac{qr}{x} + qs

  1. Rewrite in a differentiation-friendly form Express qrx\dfrac{qr}{x} as qr⋅x−1qr \cdot x^{-1} so every term is a power of xx:

f(x)=pr+psx+qrx−1+qsf(x) = pr + psx + qr x^{-1} + qs

  1. Differentiate term by term

    Apply the power rule ddx(xn)=nxn−1\dfrac{d}{dx}(x^n) = nx^{n-1} to each term:

    • ddx(pr)=0\dfrac{d}{dx}(pr) = 0 (constant)
    • ddx(psx)=ps\dfrac{d}{dx}(psx) = ps
    • ddx(qrx−1)=qr⋅(−1)x−2=−qrx2\dfrac{d}{dx}(qr x^{-1}) = qr \cdot (-1) x^{-2} = -\dfrac{qr}{x^2}
    • ddx(qs)=0\dfrac{d}{dx}(qs) = 0 (constant)
  2. Combine the results

    Sum the derivatives:

f′(x)=0+ps−qrx2+0=ps−qrx2f'(x) = 0 + ps - \dfrac{qr}{x^2} + 0 = ps - \dfrac{qr}{x^2}

Tip

When differentiating products of polynomials and simple rational terms, expansion often beats the Product Rule for speed and clarity. Save the Product Rule for when expansion would create a mess.

✓Final answer

The derivative is ps−qrx2\boxed{ps - \dfrac{qr}{x^2}}.

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