C4 January 2007 Q6
6.
| Scheme | Marks |
|---|---|
| \(y = 2^x = \mathrm{e}^{x\ln 2}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ln 2.\mathrm{e}^{x\ln 2}\) | M1 |
| Hence \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ln 2.(2^x) = 2^x\ln 2\) AG | A1 cso |
| (2) |
Notes
M1 \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ln 2.\mathrm{e}^{x\ln 2}\)
A1 cso \(2^x\ln 2\) AG
Aliter (a) Way 2
| \(\ln y = \ln(2^x)\) leads to \(\ln y = x\ln 2\) \(\dfrac{1}{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ln 2\) | M1 |
| Hence \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = y\ln 2 = 2^x\ln 2\) AG | A1 cso |
| [2] |
M1 Takes logs of both sides, then uses the power law of logarithms… … and differentiates implicitly to give \(\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x} = \ln 2\)
A1 cso \(2^x\ln 2\) AG
| Scheme | Marks |
|---|---|
| \(y = 2^{(x^2)} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x.\,2^{(x^2)}.\ln 2\) | M1 A1 |
| When \(x = 2\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(2)2^4\ln 2\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{64\ln 2} = 44.3614\ldots\) | A1 |
| (4) | |
| (6 marks) |
Notes
M1 \(Ax\,2^{(x^2)}\)
A1 \(2x.\,2^{(x^2)}.\ln 2\) or \(2x.y.\ln 2\) if \(y\) is defined
M1 Substitutes \(x = 2\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) which is of the form \(\pm k2^{(x^2)}\) or \(Ax\,2^{(x^2)}\)
A1 \(\underline{64\ln 2}\) or awrt 44.4
Aliter (b) Way 2
| \(\ln y = \ln\left(2^{x^2}\right)\) leads to \(\ln y = x^2\ln 2\) | |
| \(\dfrac{1}{y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x.\ln 2\) | M1 A1 |
| When \(x = 2\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(2)2^4\ln 2\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{64\ln 2} = 44.3614\ldots\) | A1 |
| [4] |
M1 \(\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x} = Ax.\ln 2\)
A1 \(\frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x} = 2x.\ln 2\)
M1 Substitutes \(x = 2\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) which is of the form \(\pm k2^{(x^2)}\) or \(Ax\,2^{(x^2)}\)
A1 \(\underline{64\ln 2}\) or awrt 44.4