Comprehensive assessment covering all four lessons: differentiating $e^x$, differentiating $a^x$, differentiating $\ln x$, and differentiating $\log_a x$. Includes the chain, product and quotient rules applied to exponential and logarithmic functions, gradients and tangents, second derivatives, stationary points, and real-world applications.
Assessment
Select the best answer for each question. 1 mark each.
Differentiate $y = e^x$.
Differentiate $y = e^{5x}$.
Differentiate $y = e^{-3x}$ using the chain rule.
Differentiate $y = 2^x$.
To differentiate $y = a^x$ for a general base $a > 0$, it is rewritten using the identity:
Differentiate $y = \ln x$.
Differentiate $y = \ln(5x)$.
Differentiate $y = \ln(x^2 + 1)$ using the chain rule.
Differentiate $y = \log_{10} x$.
The gradient of the tangent to $y = e^x$ at $x = 0$ is:
Differentiate $y = x e^x$ using the product rule.
Differentiate $y = x \ln x$ using the product rule.
Find $\frac{dy}{dx}$ for $y = \frac{e^x}{x}$ using the quotient rule.
If $y = e^{2x}$, then the second derivative $\frac{d^2y}{dx^2}$ equals:
The equation of the tangent to $y = \ln x$ at the point where $x = 1$ is:
Short Answer
Differentiate $y = e^{3x^2 + 1}$ using the chain rule, and evaluate $\frac{dy}{dx}$ at $x = 0$. Show all working.
Using the identity $a^x = e^{x \ln a}$, show that $\frac{d}{dx}(3^x) = 3^x \ln 3$. Show all working.
Differentiate $y = x^2 \ln x$ using the product rule, and simplify your answer. Show all working.
Find $\frac{dy}{dx}$ for $y = \frac{\ln x}{x}$ using the quotient rule. Simplify your answer. Show all working.
Find the equation of the tangent to $y = e^{2x}$ at the point where $x = 0$. Show all working.
For the curve $y = x - \ln x$ where $x > 0$, find the coordinates of the stationary point and use the second derivative to determine its nature. Show all working.
The number of bacteria in a culture is modelled by $N(t) = 500 e^{0.2t}$, where $t$ is measured in hours. Find: (a) the initial number of bacteria, and (b) the rate of growth $\frac{dN}{dt}$ at $t = 10$ hours, to the nearest whole number. Show all working.
A curve has equation $y = e^x - 3x$. Find $\frac{dy}{dx}$, determine the value of $x$ for which the curve has a stationary point, and use the second derivative to state its nature. Leave your answer in exact form. Show all working.
Q1: A, the exponential function is its own derivative, so $\frac{d}{dx}(e^x) = e^x$.
Q2: B, chain rule: $\frac{d}{dx}(e^{5x}) = e^{5x} \cdot 5 = 5e^{5x}$.
Q3: C, chain rule with inner derivative $-3$: $\frac{d}{dx}(e^{-3x}) = e^{-3x} \cdot (-3) = -3e^{-3x}$.
Q4: D, for a general base, $\frac{d}{dx}(a^x) = a^x \ln a$, so $\frac{d}{dx}(2^x) = 2^x \ln 2$.
Q5: A, since $a = e^{\ln a}$, raising to the power $x$ gives $a^x = e^{x \ln a}$, which converts any base to base $e$ for differentiation.
Q6: B, the standard result $\frac{d}{dx}(\ln x) = \frac{1}{x}$.
Q7: C, $\ln(5x) = \ln 5 + \ln x$, and $\ln 5$ is a constant, so $\frac{d}{dx}(\ln(5x)) = \frac{1}{x}$. Equivalently, the chain rule gives $\frac{5}{5x} = \frac{1}{x}$.
Q8: A, chain rule: $\frac{d}{dx}(\ln(x^2 + 1)) = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}$.
Q9: D, $\frac{d}{dx}(\log_{10} x) = \frac{1}{x \ln 10}$, using $\log_{10} x = \frac{\ln x}{\ln 10}$.
Q10: B, $y' = e^x$, so at $x = 0$ the gradient is $e^0 = 1$.
Q11: C, product rule: $(1)(e^x) + (x)(e^x) = e^x + xe^x = e^x(x + 1)$.
Q12: D, product rule: $(1)(\ln x) + (x)\left(\frac{1}{x}\right) = \ln x + 1$.
Q13: A, quotient rule: $\frac{(e^x)(x) - (e^x)(1)}{x^2} = \frac{e^x(x - 1)}{x^2}$.
Q14: B, $y' = 2e^{2x}$ and $y'' = 2 \cdot 2 e^{2x} = 4e^{2x}$.
Q15: D, at $x = 1$: $y = \ln 1 = 0$ and $y' = \frac{1}{x} = 1$. Tangent: $y - 0 = 1(x - 1)$, giving $y = x - 1$.
Q16 (3 marks): Let $u = 3x^2 + 1$, so $y = e^u$ and $\frac{du}{dx} = 6x$ [1]. By the chain rule, $\frac{dy}{dx} = e^u \cdot 6x = 6x\, e^{3x^2 + 1}$ [1]. At $x = 0$: $\frac{dy}{dx} = 6(0)\, e^{1} = 0$ [1].
Q17 (3 marks): Write $3^x = e^{x \ln 3}$ [1]. Differentiating with the chain rule, the inner derivative of $x \ln 3$ is the constant $\ln 3$, so $\frac{d}{dx}\left(e^{x \ln 3}\right) = e^{x \ln 3} \cdot \ln 3$ [1]. Since $e^{x \ln 3} = 3^x$, this equals $3^x \ln 3$, as required [1].
Q18 (3 marks): With $u = x^2$ and $v = \ln x$, $u' = 2x$ and $v' = \frac{1}{x}$ [1]. Product rule: $\frac{dy}{dx} = (2x)(\ln x) + (x^2)\left(\frac{1}{x}\right)$ [1]. Simplify: $\frac{dy}{dx} = 2x \ln x + x = x(2\ln x + 1)$ [1].
Q19 (3 marks): With $u = \ln x$ and $v = x$, $u' = \frac{1}{x}$ and $v' = 1$ [1]. Quotient rule: $\frac{dy}{dx} = \frac{\left(\frac{1}{x}\right)(x) - (\ln x)(1)}{x^2} = \frac{1 - \ln x}{x^2}$ [1]. This is already in simplest form: $\frac{dy}{dx} = \frac{1 - \ln x}{x^2}$ [1].
Q20 (4 marks): At $x = 0$: $y = e^{0} = 1$, so the point is $(0, 1)$ [1]. $y' = 2e^{2x}$, so the gradient at $x = 0$ is $2e^{0} = 2$ [1]. Tangent: $y - 1 = 2(x - 0)$ [1], which simplifies to $y = 2x + 1$ [1].
Q21 (4 marks): $y' = 1 - \frac{1}{x}$. Setting $y' = 0$ gives $\frac{1}{x} = 1$, so $x = 1$ [1]. The $y$-coordinate is $y(1) = 1 - \ln 1 = 1$, giving the stationary point $(1, 1)$ [1]. $y'' = \frac{1}{x^2}$ [1]. At $x = 1$, $y'' = 1 > 0$, so $(1, 1)$ is a local minimum [1].
Q22 (4 marks): (a) $N(0) = 500 e^{0.2(0)} = 500 e^{0} = 500$ bacteria [1]. (b) $\frac{dN}{dt} = 500 \cdot 0.2 \, e^{0.2t} = 100 e^{0.2t}$ [1]. At $t = 10$: $\frac{dN}{dt} = 100 e^{2}$ [1]. Since $e^{2} \approx 7.389$, this is approximately $739$ bacteria per hour [1].
Q23 (4 marks): $\frac{dy}{dx} = e^x - 3$ [1]. Setting $e^x - 3 = 0$ gives $e^x = 3$, so $x = \ln 3$ [1]. $y'' = e^x$ [1]. At $x = \ln 3$, $y'' = e^{\ln 3} = 3 > 0$, so the stationary point at $x = \ln 3$ is a local minimum [1].