Aussie Maths & Science Teachers: Save your time with SmarterEd
Find and simplify the rule of `f^{\prime}(x)`, where `f:R \rightarrow R, f(x)=\frac{\cos (x)}{e^x}`. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
`\frac{-(\sin x+\cos x)}{e^x}`
Using the quotient rule
| `f(x)` | `=\frac{\cos (x)}{e^x}` | |
| `f^{\prime}(x)` | `=\frac{-e^x \sin x-e^x \cos x}{e^{2 x}}` | |
| `= \frac{-e^x(\sin x+\cos x)}{e^{2 x}}` | ||
| `=\frac{-(\sin x+\cos x)}{e^x}` |
Let \(f(x)=\sin(x)e^{2x}\).
Find \(f^{'}\Big(\dfrac{\pi}{4}\Big)\). (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
\(\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \text{or}\ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}\)
\(\text{Using the product rule}\)
| \(f'(x)\) | \(=e^{2x}\cos(x)+2e^{2x}\sin(x)\) |
| \(=e^{2x}\Big(\cos(x)+2\ \sin(x)\Big)\) | |
| \(\therefore\ f’\Big(\dfrac{\pi}{4}\Big)\) | \(=e^{2(\frac{\pi}{4})}\Bigg(\cos(\dfrac{\pi}{4})+2\ \sin(\dfrac{\pi}{4})\Bigg)\) |
| \(=e^{\frac{\pi}{2}}\Bigg(\dfrac{1}{\sqrt{2}}+\sqrt{2}\Bigg)\) | |
| \(=e^{\frac{\pi}{2}}\Bigg(\dfrac{1+\sqrt2 \times \sqrt2}{\sqrt2} \Bigg) \) | |
| \(=\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \ \text{or}\ \ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}\) |
Let `f(x) = (e^x)/(cos(x))`.
Evaluate `f′(pi)`. (2 marks)
`text(See Worked Solutions)`
`f′(x) = (e^x)/(cos(x))`
| `u` | `= e^x` | `v` | `= cos(x)` |
| `u′` | `= e^x` | `v′` | `= −sin(x)` |
| `f′(x)` | `= (u′v – uv′)/(v^2)` |
| `= (e^x · cos(x) + e^x sin(x))/(cos^2(x))` |
| `f′(pi)` | `= (e^pi · cospi + e^pi sinpi)/(cos^2 pi)` |
| `= (e^pi(−1) + e^pi · 0)/((−1)^2)` | |
| `= −e^pi` |
Let `g(x) = log_e(tan(x))`. Evaluate `g′(pi/4)`. (2 marks)
`g′(pi/4) = 2`
`g(x) = log_e (tan(x))`
`text(Using Chain Rule:)`
`g′(x) = (sec^2(x))/(tan(x))`
`text(When)\ \ x = pi/4,`
| `g′(pi/4)` | `= (sec^2(pi/4))/(tan (pi/4))` |
| `=1/(1/sqrt2)^2` | |
| `=1/(1/2)` | |
| `=2` |