P7.2Differentiating standard functions

• M1 for \(x^{n} \to x^{n - 1}\) for at least one term
• A1 for two terms correct (may be unsimplified)
• A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = 5x^{4} - 5 - 2x^{-\frac{1}{2}}\), fully correct and simplified

Worked solution: In index form, \(y = x^{5} - 5x - 4x^{\frac{1}{2}}\).
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 5x^{4} - 5 - 2x^{-\frac{1}{2}} = 5x^{4} - 5 - \frac{2}{\sqrt{x}}\)

• M1 for writing \(y\) as a sum of powers of \(x\), e.g. by dividing each term of the numerator by \(2x^{2}\)
• A1 for \(y = -\frac{5}{2}x - 3x^{-1} - 3x^{-2}\)
• M1 for \(x^{n} \to x^{n - 1}\) for at least one of their terms
• A1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{5}{2} + 3x^{-2} + 6x^{-3}\), i.e. \(A = -\frac{5}{2}, \ B = 3, \ C = 6\)

Worked solution: By dividing each term of the numerator by \(2x^{2}\): \(y = -\frac{5}{2}x - 3x^{-1} - 3x^{-2}\).
So \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{5}{2} + 3x^{-2} + 6x^{-3}\), giving \(A = -\frac{5}{2}, \ B = 3, \ C = 6\).

• M1 for substituting \(x = 1\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
• A1 for \(\frac{13}{2}\) (cao)

Worked solution: At \(x = 1\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{5}{2} + 3(1)^{-2} + 6(1)^{-3} = \frac{13}{2}\).

• M1 for writing \(a^{x} = \mathrm{e}^{x\ln a}\)
• A1* for differentiating to obtain \((\ln a)\mathrm{e}^{x\ln a} = a^{x}\ln a\)

Worked solution: \(a^{x} = \mathrm{e}^{x\ln a}\), so \(\frac{\mathrm{d}}{\mathrm{d}x}\left(a^{x}\right) = (\ln a)\mathrm{e}^{x\ln a} = a^{x}\ln a\).

• M1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4^{x}\ln 4 - 6 \times 2^{x}\ln 2\)
• M1 for using \(4^{x} = (2^{x})^{2}\) and \(\ln 4 = 2\ln 2\) to obtain \(2(2^{x})^{2} - 6 \times 2^{x} = 0\) oe
• A1 for \(2^{x} = 3\)
• M1 for solving \(2^{x} = 3\) using logarithms
• A1 for \(\left(\log_{2} 3, -9\right)\)

Worked solution: \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4^{x}\ln 4 - 6 \times 2^{x}\ln 2 = 2^{x}\ln 2\left(2 \times 2^{x} - 6\right)\).
Since \(2^{x}\ln 2 \gt 0\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) when \(2^{x} = 3\), so \(x = \log_{2} 3\) \(\left(= \frac{\ln 3}{\ln 2}\right)\).
Then \(y = 3^{2} - 6 \times 3 = -9\).

• M1 for finding \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 4^{x}(\ln 4)^{2} - 6 \times 2^{x}(\ln 2)^{2}\) and substituting \(2^{x} = 3\) (or a gradient check either side)
• A1 for \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 18(\ln 2)^{2} \gt 0\), so a minimum

Worked solution: \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 4^{x}(\ln 4)^{2} - 6 \times 2^{x}(\ln 2)^{2} = (\ln 2)^{2}\left(4 \times 4^{x} - 6 \times 2^{x}\right)\).
At \(2^{x} = 3\): \((\ln 2)^{2}(36 - 18) = 18(\ln 2)^{2} \gt 0\), so the point is a minimum.