6. (a) Given that \(y = 2 ^ { x }\), and using the result \(2 ^ { x } = \mathrm { e } ^ { x \ln 2 }\), or otherwise, show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ^ { x } \ln 2\).
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Question 6:
Part (a) — Way 1:
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(y = 2^x = e^{x\ln 2}\) \(\frac{dy}{dx} = \ln 2 \cdot e^{x\ln 2}\) M1
\(\frac{dy}{dx} = \ln 2 \cdot e^{x\ln 2}\)
Hence \(\frac{dy}{dx} = \ln 2 \cdot (2^x) = 2^x \ln 2\) AG A1 cso
\(2^x \ln 2\) AG
[2]
Part (a) — Way 2 (Aliter):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\ln y = \ln(2^x)\) leads to \(\ln y = x \ln 2\) M1
Takes logs of both sides, then uses the power law of logarithms, and differentiates implicitly to give \(\frac{1}{y}\frac{dy}{dx} = \ln 2\)
\(\frac{1}{y}\frac{dy}{dx} = \ln 2\) Hence \(\frac{dy}{dx} = y\ln 2 = 2^x \ln 2\) AG A1 cso
\(2^x \ln 2\) AG
[2]
Part (b):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(y = 2^{(x^2)} \Rightarrow \frac{dy}{dx} = 2x \cdot 2^{(x^2)} \cdot \ln 2\) M1
\(A \cdot 2^{(x^2)}\)
A1 \(2x \cdot 2^{(x^2)} \cdot \ln 2\) or \(2x \cdot y \cdot \ln 2\) if \(y\) is defined
When \(x = 2\): \(\frac{dy}{dx} = 2(2) \cdot 2^4 \cdot \ln 2\) M1
Substitutes \(x = 2\) into their \(\frac{dy}{dx}\) which is of the form \(\pm k \cdot 2^{(x^2)}\) or \(A \cdot 2^{(x^2)}\)
\(\frac{dy}{dx} = 64\ln 2 = 44.3614\ldots\) A1
\(64\ln 2\) or awrt \(44.4\)
[4]
Part (b) — Way 2 (Aliter):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\ln y = \ln(2^{x^2})\) leads to \(\ln y = x^2 \ln 2\) \(\frac{1}{y}\frac{dy}{dx} = 2x \cdot \ln 2\) M1
\(\frac{1}{y}\frac{dy}{dx} = Ax \cdot \ln 2\)
A1 \(\frac{1}{y}\frac{dy}{dx} = 2x \cdot \ln 2\)
When \(x = 2\): \(\frac{dy}{dx} = 2(2) \cdot 2^4 \cdot \ln 2\) M1
Substitutes \(x = 2\) into their \(\frac{dy}{dx}\) which is of the form \(\pm k \cdot 2^{(x^2)}\) or \(A \cdot 2^{(x^2)}\)
\(\frac{dy}{dx} = 64\ln 2 = 44.3614\ldots\) A1
\(64\ln 2\) or awrt \(44.4\)
[4]
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## Question 6:
### Part (a) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2^x = e^{x\ln 2}$ | | |
| $\frac{dy}{dx} = \ln 2 \cdot e^{x\ln 2}$ | M1 | $\frac{dy}{dx} = \ln 2 \cdot e^{x\ln 2}$ |
| Hence $\frac{dy}{dx} = \ln 2 \cdot (2^x) = 2^x \ln 2$ **AG** | A1 cso | $2^x \ln 2$ **AG** |
| | **[2]** | |
### Part (a) — Way 2 (Aliter):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln y = \ln(2^x)$ leads to $\ln y = x \ln 2$ | M1 | Takes logs of both sides, then uses the power law of logarithms, and differentiates implicitly to give $\frac{1}{y}\frac{dy}{dx} = \ln 2$ |
| $\frac{1}{y}\frac{dy}{dx} = \ln 2$ | | |
| Hence $\frac{dy}{dx} = y\ln 2 = 2^x \ln 2$ **AG** | A1 cso | $2^x \ln 2$ **AG** |
| | **[2]** | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2^{(x^2)} \Rightarrow \frac{dy}{dx} = 2x \cdot 2^{(x^2)} \cdot \ln 2$ | M1 | $A \cdot 2^{(x^2)}$ |
| | A1 | $2x \cdot 2^{(x^2)} \cdot \ln 2$ or $2x \cdot y \cdot \ln 2$ if $y$ is defined |
| When $x = 2$: $\frac{dy}{dx} = 2(2) \cdot 2^4 \cdot \ln 2$ | M1 | Substitutes $x = 2$ into their $\frac{dy}{dx}$ which is of the form $\pm k \cdot 2^{(x^2)}$ or $A \cdot 2^{(x^2)}$ |
| $\frac{dy}{dx} = 64\ln 2 = 44.3614\ldots$ | A1 | $64\ln 2$ or awrt $44.4$ |
| | **[4]** | |
### Part (b) — Way 2 (Aliter):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln y = \ln(2^{x^2})$ leads to $\ln y = x^2 \ln 2$ | | |
| $\frac{1}{y}\frac{dy}{dx} = 2x \cdot \ln 2$ | M1 | $\frac{1}{y}\frac{dy}{dx} = Ax \cdot \ln 2$ |
| | A1 | $\frac{1}{y}\frac{dy}{dx} = 2x \cdot \ln 2$ |
| When $x = 2$: $\frac{dy}{dx} = 2(2) \cdot 2^4 \cdot \ln 2$ | M1 | Substitutes $x = 2$ into their $\frac{dy}{dx}$ which is of the form $\pm k \cdot 2^{(x^2)}$ or $A \cdot 2^{(x^2)}$ |
| $\frac{dy}{dx} = 64\ln 2 = 44.3614\ldots$ | A1 | $64\ln 2$ or awrt $44.4$ |
| | **[4]** | |
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6. (a) Given that $y = 2 ^ { x }$, and using the result $2 ^ { x } = \mathrm { e } ^ { x \ln 2 }$, or otherwise, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ^ { x } \ln 2$.\\
(b) Find the gradient of the curve with equation $y = 2 ^ { \left( x ^ { 2 } \right) }$ at the point with coordinates $( 2,16 )$.\\
\hfill \mbox{\textit{Edexcel C4 2007 Q6 [6]}}