Question 1
Differentiate the following using the chain rule:
(a)\(f(x) = (3x - 1)^5\)
(b)\(f(x) = (x^2 - 4)^6\)
(c)\(f(x) = (2x^3 + x)^4\)
(d)\(f(x) = \dfrac{1}{(4x+1)^3}\)
(e)\(f(x) = \sqrt{3x^2 - 5}\)
(f)\(f(x) = \dfrac{3}{\sqrt{x^2+x}}\)
Answer
\(4(6x^2+1)(2x^3+x)^3\)
Answer
\(\dfrac{3x}{\sqrt{3x^2-5}}\)
Rewrite
\(3(x^2+x)^{-1/2}\)
Answer
\(-\dfrac{3(2x+1)}{2(x^2+x)^{3/2}}\)
Question 2
Differentiate the following:
(a)\(f(x) = e^{3x}\)
(b)\(f(x) = 5e^{x^2}\)
(c)\(f(x) = 3e^{2x^2-x}\)
(d)\(f(x) = -e^{-4x}\)
(e)\(f(x) = 4e^{-x^2}\)
(f)\(f(x) = 7e^{x^3+2x-1}\)
Answer
\(3(4x-1)e^{2x^2-x}\)
Answer
\(7(3x^2+2)e^{x^3+2x-1}\)
Question 3
Differentiate the following:
(a)\(f(x) = \ln(3x)\)
(b)\(f(x) = \ln(x-4)\)
(c)\(f(x) = \ln(3x^2+1)\)
(d)\(f(x) = \ln(x^2 - x)\)
(e)\(f(x) = \ln(x^4 + 2x)\)
(f)\(f(x) = \ln(e^{2x})\)
Answer
\(\dfrac{6x}{3x^2+1}\)
Answer
\(\dfrac{2x-1}{x^2-x}\)
Answer
\(\dfrac{4x^3+2}{x^4+2x}\)
Simplify first
\(\ln(e^{2x}) = 2x\)
Question 4
Differentiate the following trig functions:
(a)\(f(x) = \sin(4x)\)
(b)\(f(x) = \cos(x^2)\)
(c)\(f(x) = \tan(3x+1)\)
(d)\(f(x) = \sin^2(x)\)
(e)\(f(x) = 2\cos(x^3 - x)\)
(f)\(f(x) = \sqrt{\sin x}\)
Rewrite as composite
\((\sin x)^2\)
Answer
\(2\sin x \cos x = \sin 2x\)
Answer
\(-2(3x^2-1)\sin(x^3-x)\)
Rewrite
\((\sin x)^{1/2}\)
Answer
\(\dfrac{\cos x}{2\sqrt{\sin x}}\)
Question 5
Find the equation of the tangent to \(f(x) = (2x^2 - 1)^3\) at \(x = 1\). Leave your answer in the form \(y = mx + c\).
Find gradient
\(f'(x) = 3(2x^2-1)^2 \cdot 4x = 12x(2x^2-1)^2\)
At \(x=1\)
\(f'(1) = 12(1)(1)^2 = 12,\quad f(1) = 1\)
Question 6
\(f(x) = e^{x^2 - 4x}\). Show that \(f'(x) = 0\) when \(x = 2\) and find the value of \(f(2)\). Leave your answer in exact form.
Differentiate
\(f'(x) = (2x-4)e^{x^2-4x}\)
Set to zero
\((2x-4)e^{x^2-4x}=0\). Since \(e^{x^2-4x}>0\) for all \(x\), we need \(2x-4=0 \Rightarrow x=2\) ✓
Answer
\(f(2) = e^{4-8} = e^{-4}\)
Question 7
[Non GDC] The graph of \(f(x) = \ln(x^2 - 3x + 3)\) is shown below.
(a)Find \(f'(x)\).
(b)Show that the gradient is \(0\) when \(x = \dfrac{3}{2}\) and find \(f\!\left(\dfrac{3}{2}\right)\).
(c)Find the coordinates of the minimum point. Leave your answer in exact form.
Answer
\(f'(x) = \dfrac{2x-3}{x^2-3x+3}\)
Set numerator to zero
\(2x-3=0 \Rightarrow x=\tfrac{3}{2}\) ✓
Answer
\(f\!\left(\tfrac{3}{2}\right) = \ln\!\left(\tfrac{9}{4}-\tfrac{9}{2}+3\right) = \ln\!\left(\tfrac{3}{4}\right)\)
Answer
Minimum at \(\left(\dfrac{3}{2},\ \ln\dfrac{3}{4}\right)\)