Differentiate with respect to \(x\) :
\(f(x)=\log _e\left(\dfrac{x^3}{3-2 x}\right)\) (3 marks)
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Differentiate with respect to \(x\) :
\(f(x)=\log _e\left(\dfrac{x^3}{3-2 x}\right)\) (3 marks)
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\(f^{\prime}(x)=\dfrac{9-4 x}{x(3-2 x)}\)
\begin{align}
\begin{aligned}
\text {Let} \ \ & u=x^3 & u^{\prime}=3 x^2 \\
& v=3-2 x & v^{\prime}=-2
\end{aligned}
\end{align}
| \(f^{\prime}(x)\) | \(=\dfrac{\dfrac{v u^{\prime}-u v^{\prime}}{v^2}}{\frac{x^3}{3-2 x}}\) |
| \(=\dfrac{(3-2 x) 3 x^2-x^3(-2)}{(3-2 x)^2} \times \dfrac{3-2 x}{x^3}\) | |
| \(=\dfrac{x^2(9-6 x+2 x)}{(3-2 x)} \times \dfrac{1}{x^3}\) | |
| \(=\dfrac{9-4 x}{x(3-2 x)}\) |
Given the function \(f(x)=\log _{10} x^x\), which of the following expressions is equal to \(f^{\prime}(x)\) ?
\(B\)
\(f(x)=\log _{10} x^x=x \log _{10} x\)
\(\text{Using product rule:}\)
| \(f^{\prime}(x)\) | \(=x \cdot \dfrac{1}{x \cdot \ln 10}+1 \cdot \log _{10} x\) |
| \(=\dfrac{1}{\ln 10}+\log _{10} x\) | |
| \(=\dfrac{1}{\ln 10}+\dfrac{\ln x}{\ln 10}\) | |
| \(=\dfrac{\ln x+1}{\ln 10}\) |
\(\Rightarrow B\)
Let \(f(x)=\log _e\left(x^3-3 x+2\right)\). Find \(f^{\prime}(3)\) (2 marks) --- 6 WORK AREA LINES (style=lined) --- \(\dfrac{6}{5}\)
\(f(x)\)
\(=\log_{e}(x^3-3x+2)\)
\(f^{\prime}(x)\)
\(=\dfrac{3x^2-3}{x^3-3x+2}\)
\(f^{\prime}(3)\)
\(=\dfrac{3(3)^2-3}{(3)^3-3(3)+2}=\dfrac{6}{5}\)
Let \(y=e^x \cos\,3 x\). Find \(\dfrac{d y}{d x}\) (2 marks) \(e^x (\cos(3x)-3\sin(3x))\)
\(y\)
\(=e^x \cos(3x)\)
\(\dfrac{dy}{dx}\)
\(=e^x.(-3\sin(3x))+\cos(3x).e^x\)
\(=e^x(\cos(3x)-3\sin(3x))\)
Let \(y=\dfrac{x^2-x}{e^x}\).
Find and simplify \(\dfrac{dy}{dx}\). (2 marks)
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\(\dfrac{-x^2+3x-1}{e^x}\)
\(\text{Using the quotient rule:}\)
| \(\dfrac{dy}{dx}\) | \(=\dfrac{e^x(2x-1)-(x^2-x)e^x}{(e^x)^2}\) |
| \(=\dfrac{e^x(-x^2+3x-1)}{e^{2x}}\) | |
| \(=\dfrac{-x^2+3x-1}{e^x}\) |
Differentiate `pi^(2x)`. (2 marks)
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`2log_e(pi) * pi^(2x)`
| `y` | `=pi^(2x)` |
| `dy/dx` | `=log_e(pi) * 2 * pi^(2x)` |
| `=2log_e(pi) *pi^(2x)` |
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)
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`-6e^(-3x)`
| `y` | `=2e^(-3x)` |
| `dy/dx` | `=-3 xx 2e^(-3x)=-6e^(-3x)` |
Differentiate with respect to `x`:
`e^(tan(2x))` (2 marks)
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`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))` |
If `f(x)=log_2(x^(2x))`, which expression is equal to `f^(′)(x)`?
`B`
| `f(x)` | `=log_2(x^(2x))` |
| `=2x log_2x` | |
| `=(2x lnx)/ln2` |
| `f^(′)(x)` | `=1/ln2 (2x*1/x + 2lnx)` |
| `=2/ln2 + (2lnx)/ln2` | |
| `=2/ln2 + 2log_2x` |
`=> B`
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)` |
`:. \ text{when}\ x=5,\ \ dy/dx=log_e 5 +2`
Let `f(x) = (e^x)/((x^2-3))`.
Find `f^{′}(x)`. (2 marks)
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`{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 with respect to `x`:
`log_e x^x`. (2 marks)
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`1 + log_ex`
| `y` | `=log_e x^x=xlog_ex` |
| `dy/dx` | `=x*1/x + log_ex` |
| `=1 + log_ex` |
Differentiate `5^(x^2)5x`. (2 marks)
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`5^(x^2 + 1)(ln5*2x^2 + 1)`
| `y` | `= 5^(x^2) * 5x` |
| `(dy)/(dx)` | `= ln5*2x*5^(x^2)*5x + 5^(x^2)*5` |
| `=5^(x^2)(ln5*10x^2 + 5)` | |
| `=5^(x^2 + 1)(ln5*2x^2 + 1)` |
Differentiate `3x 6^x`. (2 marks)
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`3*6^x(xln6 +1)`
`y= 3x * 6^x`
`text{Using the product rule:}`
`(dy)/(dx)= 3*6^x + ln6 * 6^x *3x= 3*6^x(1 + xln6)`
Differentiate with respect to `x`:
`10^(5x^2-3x)`. (2 marks)
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`(dy)/(dx) = ln 10 (10x-3) * 10^(5x^2-3x)`
`y = 10^(5x^2-3x)`
`(dy)/(dx) = ln 10 (10x-3) * 10^(5x^2-3x)`
Differentiate `log_2 x^2` with respect to `x`. (2 marks)
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`2/(xln2)`
| `y` | `= log_2 x^2` |
| `(dy)/(dx)` | `= {:d/(dx):} ((lnx^2)/(ln2))` |
| `= 1/(ln2) · d/(dx)(ln x^2)` | |
| `= 1/(ln2) · (2x)/(x^2)` | |
| `= 2/(xln2)` |
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 `sin(ln x),` where `x > 0`?
`D`
| `y` | `= sin (ln x)` |
| `(dy)/(dx)` | `= cos (ln x) xx d/(dx) (ln x)` |
| `= cos (ln x) xx 1/x` | |
| `= (cos (ln x))/x` |
`=> D`
Differentiate `x^3 ln x`. (2 marks)
`x^2 (3 ln\ x + 1)`
`y = x^3 ln\ x`
`text(Using the product rule:)`
| `(dy)/(dx)` | `= 3x^2 * ln\ x + x^3 * 1/x` |
| `= x^2 (3 ln\ x + 1)` |
What is the derivative of `e^(x^2)`?
`C`
| `y` | `= e^(x^2)` |
| `(dy)/(dx)` | `= 2x e^(x^2)` |
`=> C`
What is the derivative of `ln (cos x)?`
`B`
| `y` | `= ln (cos x)` |
| `(dy)/(dx)` | `= (-sin x)/(cos x)` |
| `= -tan x` |
`=> B`
Differentiate `y = (x + 4) ln\ x`. (2 marks)
`ln\x + 4/x +1`
`y = (x + 4) ln\ x`
`text(Using the product rule)`
| `(dy)/(dx)` | `= d/(dx) (x + 4) * ln x + (x + 4) d/(dx) ln\ x` |
| `= ln x + (x + 4) 1/x` | |
| `= ln x + 4/x + 1` |
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` |
Find the equation of the tangent to `y = log_ex` at the point `(e, 1)`. (2 marks)
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`y = x/e`
`y = log_ex`
`dy/dx = 1/x`
`text(At)\ \ (e\ ,\ 1),\ \ m = 1/e`
`text(Equation of tangent,)\ \ m = 1/e,\ text(through)\ (e\ ,\ 1)`
| `y-y_1` | `= m(x-x_1)` |
| `y-1` | `= 1/e(x-e)` |
| `y-1` | `= x/e-1` |
| `y` | `= x/e` |
Differentiate with respect to `x`:
`x^2 log_e x` (2 marks)
`x + 2x log_e x`
| `y` | `= x^2 log_e x` |
| `dy/dx` | `= x^2 * 1/x + 2x * log_e x` |
| `= x + 2x log_e x` |
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 `ln(5x+2)` with respect to `x`. (2 marks)
`5/(5x+2)`
| `y` | `=ln(5x+2)` |
| `dy/dx` | `=5/(5x+2)` |
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 with respect to `x`
`(x-1)log_ex` (2 marks)
`log_ex+1-1/x`
| `y` | `=(x-1)log_ex` |
| `dy/dx` | `=1(log_ex)+(x-1)1/x` |
| `=log_ex+1-1/x` |
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)` |