Hint
Exclude the zero of the denominator. Use the quotient rule and \(\displaystyle e^{0}=1\).
Worked solution
(a)
\(\displaystyle \frac{15 e^{3 x}}{x + 3} - \frac{5 e^{3 x}}{\left(x + 3\right)^{2}}\) follows by differentiating each factor or inner function as required.
Answer: \(\displaystyle f'(x) = \frac{15 e^{3 x}}{x + 3} - \frac{5 e^{3 x}}{\left(x + 3\right)^{2}}\)
(b)
Differentiate the complete first derivative and simplify. \(\displaystyle \frac{5 \left(9 - \frac{6}{x + 3} + \frac{2}{\left(x + 3\right)^{2}}\right) e^{3 x}}{x + 3}\)
Answer: \(\displaystyle \frac{5 \left(9 - \frac{6}{x + 3} + \frac{2}{\left(x + 3\right)^{2}}\right) e^{3 x}}{x + 3}\)
(c)
The point has height \(\displaystyle \frac{5}{3}\) and gradient \(\displaystyle \frac{40}{9}\). Use \(y-y_0=m(x-x_0)\) to obtain \(\displaystyle y=\frac{40 x}{9} + \frac{5}{3}\).
Answer: \(\displaystyle y=\frac{40 x}{9} + \frac{5}{3}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Select and apply the differentiation rule, retaining every inner derivative.
- 1 mark: Obtain an algebraically equivalent complete derivative.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not solve \(\displaystyle e^{kx}=0\).
(b) · 2 marks
- 1 mark: Differentiate the first derivative consistently.
- 1 mark: Obtain the complete correct second derivative.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: A stationary point must lie on the original curve.
(c) · 2 marks
- 1 mark: Evaluate both the ordinate and derivative at the specified abscissa.
- 1 mark: Give a complete correct tangent equation.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not solve \(\displaystyle e^{kx}=0\).
Total: 6 marks.