Let `y= (x + 5) log_e (x)`.
Find `(dy)/(dx)` when `x = 5`. (2 marks)
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Let `y= (x + 5) log_e (x)`.
Find `(dy)/(dx)` when `x = 5`. (2 marks)
`log_e 5 +2`
`(dy)/(dx)` | `= 1 xx log_e x + (x + 5) * (1)/(x)` |
`= log_e x + (x + 5)/(x)` |
`:. dy/dx|_(x=5)=log_e 5 +2`
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 with respect to `x`
`(x - 1)log_e x` (2 marks)
`log_e x + 1 – 1/x`
`y` | `= (x – 1) log_e x` |
`(dy)/(dx)` | `= 1(log_e x) + (x – 1) 1/x` |
`= log_e x + 1 – 1/x` |
Differentiate with respect to `x`:
`x^2log_ex` (2 marks)
`x + 2xlog_ex`
`y` | `= x^2 log_e x` |
`(dy)/(dx)` | `= x^2 · 1/x + 2x · log_ex` |
`= x + 2xlog_ex` |
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 log_e (x)` with respect to `x.` (2 marks)
`log_e (x) + 1`
`text(Using Product rule:)`
`(fg)′` | `= f′g + fg′` |
`d/(dx) (x log_e (x))` | `= 1 xx log_e (x) + x (1/x)` |
`= log_e (x) + 1` |
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)` |
If `y = x^2 log_e (x)`, find `(dy)/(dx)`. (2 marks)
`2x log_e (x) + x`
`text(Using Product Rule:)`
`(fg)′` | `= f prime g + f g prime` |
`(dy)/(dx)` | `= 2x log_e (x) + x^2 (1/x)` |
`= 2x log_e (x) + x` |