Detailed solution to Problem Set 15 Exercise 7

Question:

Find d(4 + 4∕x).

Solutions:

You can use the Quotient Rule (d(u∕v) = (v du − u dv)∕v²): d(4 + 4∕x) = d(4∕x) = (x d(4) − 4 dx)∕x² = (x ⋅ 0 − 4 dx)∕x² = −4 dx∕x², which you can write as (−4∕x²) dx or −4x⁻² dx if you want.

Or you can avoid the Quotient Rule by treating 4∕x as 4x⁻¹: d(4 + 4x⁻¹) = d(4x⁻¹) = 4 d(x⁻¹) = 4(−1x⁻¹⁻¹ dx) = −4x⁻² dx, which you can write as (−4∕x²) dx or −4 dx∕x² if you want.

Or you can use the relatively obscure Reciprocal Rule (d(∕u) = −du∕u², where ∕a indicates the reciprocal of a so that b∕a can be read as b ⋅ ∕a): d(4 + 4∕x) = d(4∕x) = 4 d(∕x) = 4(−dx∕x²) = −4 dx∕x², which again you can write as (−4∕x²) dx or −4x⁻² dx if you want.

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