Question 1
Differentiate the following functions (simplify first if required):
(a)\(f(x) = 3e^x\)
(b)\(f(x) = -5e^x\)
(c)\(f(x) = \ln(x)\)
(d)\(f(x) = 4\ln(x) - 3e^x\)
(e)\(f(x) = \dfrac{3e^{3x} - e^x}{e^x}\)
(f)\(f(x) = \ln(3x)\)
Answer
\(\dfrac{4}{x} - 3e^x\)
Simplify
\(f(x) = 3e^{2x} - 1\)
Log law
\(\ln(3x) = \ln 3 + \ln x\)
Question 2
Differentiate the following functions using the product rule:
(a)\(f(x) = x^2 e^x\)
(b)\(f(x) = e^{2x}\ln x\)
(c)\(f(x) = (\ln x)^3\)
Product rule: \(u=x^2,\;v=e^x\)
\(u'=2x,\;v'=e^x\)
Answer
\(e^x(x^2 + 2x) = xe^x(x+2)\)
Product rule: \(u=e^{2x},\;v=\ln x\)
\(u'=2e^{2x},\;v'=\tfrac{1}{x}\)
Answer
\(2e^{2x}\ln x + \dfrac{e^{2x}}{x} = e^{2x}\!\left(2\ln x + \dfrac{1}{x}\right)\)
Chain rule
\((\ln x)^3\); let \(u = \ln x\)
Answer
\(\dfrac{3(\ln x)^2}{x}\)
Question 3
Find the equation of the tangent to \(f(x) = x^2 \ln x\) at \(x = e\). Leave your answer in the form \(y = mx + c\).
Product rule
\(f'(x) = 2x\ln x + x^2 \cdot \tfrac{1}{x} = 2x\ln x + x\)
At \(x=e\)
\(f'(e) = 2e\cdot 1 + e = 3e,\quad f(e) = e^2\)
Question 4
Find the equation of the tangent to \(f(x) = 3x^2 e^{-x}\) at \(x = 1\). Leave your answer in terms of \(e\).
Product rule: \(u=3x^2,\;v=e^{-x}\)
\(f'(x) = 6xe^{-x} - 3x^2 e^{-x} = 3xe^{-x}(2-x)\)
At \(x=1\)
\(f'(1) = 3e^{-1}(1) = \tfrac{3}{e},\quad f(1) = 3e^{-1} = \tfrac{3}{e}\)
Answer
\(y = \dfrac{3}{e}x\)
Question 5
Differentiate the following using the quotient rule where appropriate:
(a)\(y = \dfrac{x}{2+x}\)
(b)\(y = \dfrac{x^2 + 3x}{x - 2}\)
(c)\(y = \dfrac{5x - 1}{3x + 2}\)
Quotient rule: \(u=x,\;v=2+x\)
\(u'=1,\;v'=1\)
Answer
\(\dfrac{(2+x) - x}{(2+x)^2} = \dfrac{2}{(2+x)^2}\)
Quotient rule
\(u=x^2+3x,\;v=x-2,\;u'=2x+3,\;v'=1\)
Answer
\(\dfrac{(2x+3)(x-2)-(x^2+3x)}{(x-2)^2} = \dfrac{x^2-4x-6}{(x-2)^2}\)
Answer
\(\dfrac{5(3x+2)-3(5x-1)}{(3x+2)^2} = \dfrac{13}{(3x+2)^2}\)
Question 6
Differentiate the following using the quotient rule:
(a)\(y = \dfrac{3x^2-1}{2x+3}\)
(b)\(y = \dfrac{e^x}{x+2}\)
(c)\(y = \dfrac{2x+5}{e^x}\)
Answer
\(\dfrac{6x(2x+3)-2(3x^2-1)}{(2x+3)^2} = \dfrac{6x^2+18x+2}{(2x+3)^2}\)
Answer
\(\dfrac{e^x(x+2)-e^x}{(x+2)^2} = \dfrac{e^x(x+1)}{(x+2)^2}\)
Answer
\(\dfrac{2e^x-(2x+5)e^x}{e^{2x}} = \dfrac{-2x-3}{e^x}\)
Question 7
[Non GDC] The graph of \(f(x) = x^2 e^{-x}\) is shown below.
(a)Find \(f'(x)\).
(b)Show that the gradient is \(0\) when \(x = 0\) and \(x = 2\).
(c)Find the coordinates of the local maximum.
Product rule
\(u=x^2,\;v=e^{-x}\)
Answer
\(f'(x) = 2xe^{-x} - x^2e^{-x} = xe^{-x}(2-x)\)
Set to zero
\(xe^{-x}(2-x)=0\). Since \(e^{-x}>0\): \(x=0\) or \(x=2\) ✓
Test sign of \(f'(x)\) around \(x=2\)
For \(x<2\): \(f'>0\); for \(x>2\): \(f'<0\) → local max at \(x=2\)
Answer
Local maximum at \(\left(2,\; 4e^{-2}\right)\)