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QLDMath MethodsQuick questions
Unit 3: Further calculus and statistics
Quick questions on Derivatives of exponential and logarithmic functions (QCE Mathematical Methods Unit 3)
14short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.
What is the base case?Show answer
The exponential function with base $e$ is its own derivative.
What is chain-rule generalisations?Show answer
For any differentiable inner function $f(x)$:
What is other bases?Show answer
For $a > 0$ and $a \neq 1$, use the change of base $a^x = e^{x \ln a}$, which gives
What is logarithm laws first?Show answer
Before differentiating a complicated logarithm, simplify using logarithm laws. For example,
What is direct chain rule?Show answer
Let $f(x) = -2x$, so $f'(x) = -2$. Then $\dfrac{dy}{dx} = -2 e^{-2x}$.
What is logarithm of a polynomial?Show answer
Differentiate $y = \ln(x^2 + 4)$.
What is non-$e$ exponential?Show answer
$\dfrac{dy}{dx} = (\ln 5) \cdot 5^x \approx 1.6094 \cdot 5^x$.
What is logarithm laws save work?Show answer
Differentiate $y = \ln\!\left( \dfrac{x^3}{x^2 + 1} \right)$.
What is modelling context?Show answer
A quantity decays according to $Q(t) = 200 e^{-0.05 t}$ milligrams, with $t$ in days. The rate of decay at $t = 10$ is
What is treating $a^x$ like $x^n$?Show answer
$\frac{d}{dx}(3^x) \neq x \cdot 3^{x-1}$. The correct answer is $(\ln 3) \cdot 3^x$.
What is forgetting the chain rule factor?Show answer
$\frac{d}{dx}(e^{2x}) = 2 e^{2x}$, not $e^{2x}$. The factor of $2$ comes from $f'(x) = 2$.
What is missing the domain on $\ln x$?Show answer
The natural log is only defined for $x > 0$. If a domain crosses zero, you may need $\ln|x|$ instead. In Paper 1, state the domain when QCAA asks for it.
What is skipping logarithm laws?Show answer
Differentiating $\ln(uv)$ directly with the chain rule still works but is slower and more error-prone than splitting first to $\ln u + \ln v$.
What is forgetting the sign in decay?Show answer
A decay model $Q(t) = A e^{-kt}$ with $k > 0$ has derivative $-k A e^{-kt}$, which is negative. The negative sign carries through and is part of the answer.