Let `f(x) = e^(x^2)`.
Find `f prime (3)`. (3 marks)
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Let `f(x) = e^(x^2)`.
Find `f prime (3)`. (3 marks)
`6e^9`
`text(Using Chain Rule:)`
`f prime (x)` | `= 2xe^(x^2)` |
`f prime (3)` | `= 2 (3) e^((3)^2)` |
`= 6e^9` |
Evaluate `f′(1)`, where `f: R -> R, \ f(x) = e^(x^2 - x + 3)`. (2 marks)
`e^3`
`f(x)` | `= e^(x^2 – x + 3)` |
`f′(x)` | `= (2x – 1)e^(x^2 – x + 3)` |
`f′(1)` | `= (2 – 1)e^(1 – 1 + 3)` |
`= e^3` |
Let `f(x) = (e^x)/((x^2 - 3))`.
Find `f′(x)`. (2 marks)
`{e^x(x^2 – 2x – 3)}/{(x^2 – 3)^2}`
`text(Let) \ \ u = e^x \ \ => \ \ u′ = e^x`
`v = (x^2 – 3) \ \ => \ \ v′ = 2x`
`f′(x)` | `= {e^x(x^2 – 3) – 2x e^x}/{(x^2 – 3)^2}` |
`= {e^x(x^2 – 2x – 3)}/{(x^2 – 3)^2}` |
Let `y = (2e^(2x) - 1)/e^x`.
Find `(dy)/(dx)`. (2 marks)
`(dy)/(dx) = 2e^x + e^(-x)`
`text(Method 1)`
`y` | `= 2e^x – e^(-x)` |
`(dy)/(dx)` | `= 2e^x + e^(-x)` |
`text(Method 2)`
`(dy)/(dx)` | `= (4e^(2x) ⋅ e^x – (2e^(2x) – 1) e^x)/(e^x)^2` |
`= (4e^(3x) – 2e^(3x) + e^x)/e^(2x) ` | |
`= (2e^(2x) + 1)/e^x` |
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 `f(x) = xe^(3x)`. Evaluate `f′(0)`. (3 marks)
`1`
`text(Using Product Rule:)`
`(gh)′ = g′h + gh′`
`f′(x)` | `= x(3e^(3x)) + 1 xx e^(3x)` |
`:.f′(0)` | `= 0 + e^0` |
`= 1` |
Differentiate `(e^x + x)^5`. (2 marks)
`5(e^x + 1)(e^x + x)^4`
`y` | `= (e^5 + x)^5` |
`(dy)/(dx)` | `= 5(e^x + x)^4 xx d/(dx)(e^x + x)` |
`= 5(e^x + x)^4 xx (e^x + 1)` | |
`= 5(e^x + 1)(e^x + x)^4` |
Differentiate with respect to `x`:
`(2x)/(e^x + 1)` (2 marks)
`(2(e^x + 1 – xe^x))/((e^x + 1)^2)`
`y = (2x)/(e^x + 1)`
`u` | `= 2x` | `v` | `= e^x + 1` |
`u′` | `= 2` | `v′` | `= e^x` |
`(dy)/(dx)` | `= (u′v – uv′)/(v^2)` |
`= (2(e^x + 1) – 2x(e^x))/((e^x + 1)^2)` | |
`= (2e^x + 2 – 2x · e^x)/((e^x + 1)^2)` | |
`= (2(e^x + 1 – xe^x))/((e^x + 1)^2)` |
Differentiate `(e^x + 1)^2` with respect to `x`. (2 marks)
`2e^x(e^x + 1)`
`y` | `= (e^x + 1)^2` | |
`(dy)/(dx)` | `= 2(e^x + 1)^1 xx d/(dx) (e^x + 1)` | |
`= 2e^x(e^x + 1)` |
Let `f(x) = x^2e^(5x)`.
Evaluate `f′(1)`. (2 marks)
`7e^5`
`text(Using Product Rule:)`
`(fg)′` | `= f′g + fg′` |
`f′(x)` | `= 2xe^(5x) + 5x^2 e^(5x)` |
`f′(1)` | `= 2(1)e^(5(1)) + 5(1)^2 e^(5(1))` |
`= 7e^5` |
For `y = e^(2x) cos (3x)` the rate of change of `y` with respect to `x` when `x = 0` is
`B`
`y` | `= e^(2x) cos (3x)` |
`dy/dx` | `=e^(2x) xx -3sin(3x) + 2e^(2x) xx cos (3x)` |
`=e^(2x)(-3sin(3x) + 2cos(3x))` |
`text(When)\ \ x = 0,`
`dy/dx= 2`
`=> B`
Differentiate `x^3 e^(2x)` with respect to `x`. (2 marks)
`3x^2 e^(2x) + 2x^3 e^(2x)`
`text(Using Product rule:)`
`(fg)′` | `= f′g + fg′` |
`d/(dx) (x^3 e^(2x))` | `= 3x^2 e^(2x) + 2x^3 e^(2x)` |