| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Show that derivative equals expression |
| Difficulty | Moderate -0.3 This is a straightforward algebraic manipulation followed by standard differentiation. Part (a) requires expanding and simplifying to match coefficients (routine algebra), while part (b) involves differentiating three power terms and simplifying to the given form. The question is slightly easier than average because it's purely procedural with no problem-solving required, though the algebraic manipulation and simplification require care. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.07i Differentiate x^n: for rational n and sums1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(f(x) = \frac{x^2 - 8x + 16}{2x^3}\) | M1 |
| \(f(x) = \frac{1}{2}x^{-1} - 4x^{-1} + 8x^{-1},\) \(A = \frac{1}{2}, B = -4, C = 8\) | A2 | |
| (b) | \(f'(x) = \frac{1}{4}x^{\frac{1}{3}} - 2x^{-1} - 4x^{-1}\) | M1 A2 |
| \(f'(x) = \frac{1}{4}x^{-\frac{2}{3}}(3x^2 - 8x - 16)\) | M1 | |
| \(f'(x) = \frac{1}{4}x^{-\frac{2}{3}}(3x + 4)(x - 4) = \frac{(3x+4)(x-4)}{4x^{\frac{2}{3}}}\) | M1 A1 | (9) |
(a) | $f(x) = \frac{x^2 - 8x + 16}{2x^3}$ | M1 |
| $f(x) = \frac{1}{2}x^{-1} - 4x^{-1} + 8x^{-1},$ $A = \frac{1}{2}, B = -4, C = 8$ | A2 |
(b) | $f'(x) = \frac{1}{4}x^{\frac{1}{3}} - 2x^{-1} - 4x^{-1}$ | M1 A2 |
| $f'(x) = \frac{1}{4}x^{-\frac{2}{3}}(3x^2 - 8x - 16)$ | M1 |
| $f'(x) = \frac{1}{4}x^{-\frac{2}{3}}(3x + 4)(x - 4) = \frac{(3x+4)(x-4)}{4x^{\frac{2}{3}}}$ | M1 A1 | (9)$$\text{f}(x) = \frac{(x-4)^2}{2x^{\frac{1}{2}}}, \quad x > 0.$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $A$, $B$ and $C$ such that
$$\text{f}(x) = Ax^{\frac{3}{2}} + Bx^{\frac{1}{2}} + Cx^{-\frac{1}{2}}.$$ [3]
\item Show that
$$\text{f}'(x) = \frac{(3x+4)(x-4)}{4x^{\frac{3}{2}}}.$$ [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q7 [9]}}