Let \(y=\dfrac{x^2-x}{e^x}\).
Find and simplify \(\dfrac{dy}{dx}\). (2 marks)
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Let \(y=\dfrac{x^2-x}{e^x}\).
Find and simplify \(\dfrac{dy}{dx}\). (2 marks)
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\(\dfrac{-x^2+3x-1}{e^x}\)
\(\text{Using the quotient rule:}\)
\(\dfrac{dy}{dx}\) | \(=\dfrac{e^x(2x-1)-(x^2-x)e^x}{(e^x)^2}\) |
\(=\dfrac{e^x(-x^2+3x-1)}{e^{2x}}\) | |
\(=\dfrac{-x^2+3x-1}{e^x}\) |
If `f(x)=e^(g(x^(2)))`, where `g` is a differentiable function, then `f^(′)(x)` is equal to
`C`
`f(x)=e^(g(x^2))`
`text{Using the chain rule (twice):}`
`f^(′)(x)` | `=d/dx[g(x^2)] * e^(g(x^2))` | |
`=2x*g^(′)(x^2)*e^(g(x^2))` |
`=> C`
Differentiate `y = 2e^(−3x)` with respect to `x`. (2 mark)
`-6e^(-3x)`
`y` | `=2e^(-3x)` | |
`dy/dx` | `=-3 xx 2e^(-3x)` | |
`=-6e^(-3x)` |
Differentiate with respect to `x`:
`e^(tan(2x))` (2 marks)
`2 sec^2(2x)* e^(tan(2x))`
`y` | `=e^(tan(2x))` |
`dy/dx` | `= d/(dx)tan(2x) xx e^(tan(2x))` |
`= 2 sec^2(2x)* e^(tan(2x))` |
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)
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`(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` |
Differentiate `e^x/(x + 1)`. (2 marks)
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`(xe^x)/(x + 1)^2`
`y = e^x/(x + 1)`
`text(Differentiate using quotient rule:)`
`u = e^x` | `v = x + 1` |
`u prime = e^x` | `v prime = 1` |
`(dy)/(dx)` | `= (u prime v – u v prime)/v^2` |
`= (e^x(x + 1) – e^x ⋅ 1)/(x + 1)^2` | |
`= (x e^x)/(x + 1)^2` |
What is the derivative of `e^(x^2)`?
`C`
`y` | `= e^(x^2)` |
`(dy)/(dx)` | `= 2x e^(x^2)` |
`=> C`
Differentiate `(e^x + x)^5`. (2 marks)
`5 (e^x + 1) (e^x + x)^4`
`y` | `= (e^x + 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)
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`{2(e^x + 1 – xe^x)}/(e^x + 1)^2`
`y = (2x)/(e^x + 1)`
`u` | `= 2x` | `\ \ \ \ \v` | `= e^x + 1` |
`u prime` | `= 2` | `\ \ \ \ \ v prime` | `= e^x` |
`(dy)/(dx)` | `= (u prime v – uv prime)/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 with respect to `x`.
`(e^x+1)^2`. (2 marks)
`2e^x(e^x+1)`
`y` | `=(e^x+1)^2` |
`dy/dx` | `=2(e^x+1)^1xxd/(dx) (e^x+1)` |
`=2e^x(e^x+1)` |
Differentiate `(3+e^(2x))^5`. (2 marks)
`10e^(2x)(3+e^(2x))^4`
`y=(3+e^(2x))^5`
`(dy)/dx` | `=5(3+e^(2x))^4 xx d/(dx)(3+e^(2x))` |
`=5(3+e^(2x))^4 xx 2e^(2x)` | |
`=10e^(2x)(3+e^(2x))^4` |
Differentiate `x^2e^x` (2 marks)
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`xe^x(x+2)`
`text{Using the product rule}`
`text(Let)\ \ u=x^2,` | `\ \ \ \ \ \ u^{\prime}=2x` |
`text(Let)\ \ v=e^x,` | `\ \ \ \ \ \ v^{\prime}=e^x` |
`{d(uv)}/dx` | `=u prime v+v prime u` |
`=2x e^x +x^2 e^x ` | |
`=xe^x(x+2)` |