HOW DO YOU FIND THE COEFFICIENT OF x6 IN THE DERIVATIVE OF (1 + 2x)7?
Differentiate first, then pick the term. Doing it the other way round means expanding seven terms you will throw away, and in an ESAT Mathematics 2 module you have about 89 seconds.
Given y = (1 + 2x)7, what is the coefficient of x6 in dy⁄dx?
Differentiate the whole bracket with the chain rule before you expand anything. You then want the x⁶ term of a sixth power, which is one binomial coefficient, not a whole expansion.
- dy⁄dx = 7(1 + 2x)6 × 2 = 14(1 + 2x)6
- the x6 term of (1 + 2x)6 is C(6,6)(2x)6 = 64x6
- 14 × 64 = 896
448 is what you get if you forget the inner derivative and differentiate (1 + 2x)⁷ as 7(1 + 2x)⁶, dropping the factor of 2. It is the most common single error on chain-rule binomials and it produces an option that looks entirely reasonable.
2688 is what you get if you take the wrong term of the expansion using C(6,5)(2x)⁵ instead of C(6,6)(2x)⁶. A sixth power has seven terms, and the one you want is the last of them, not the second to last.
When the power you want equals the power of the bracket, the binomial coefficient is 1 and the whole thing collapses to the outer factor times the inner coefficient raised to that power. Here that is 14 × 2⁶ = 896, which you can do in your head. No expansion, no Pascal’s triangle.
THE GENERAL RULE
Whenever a question asks for one coefficient rather than an expansion, you are being tested on whether you can isolate a single term. Expanding is not wrong, it is just slow enough to cost you two other questions, and in a module that gives you roughly 89 seconds each that is the whole difference. Differentiate first, simplify the bracket, then take the one term you need.
Updated · written to the published ESAT content specification