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9709 · 2.4

Differentiation flashcards

Revision flashcards for Cambridge 9709 Differentiation (syllabus 2.4). Flip, recall, then mark a real past-paper question.

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    What is the Product Rule for differentiation?

    If $y = uv$, where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$. In words: 'first times derivative of the second, plus second times derivative of the first'.

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    What is the Quotient Rule for differentiation?

    If $y = \frac{u}{v}$, where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. A common mnemonic is 'low d-high minus high d-low, over low-low'.

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    What is the condition for a point to be a stationary point?

    A point $(x, y)$ on the curve $y = f(x)$ is a stationary point if $f'(x) = 0$.

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    How is the second derivative, $f''(x)$, used to classify stationary points?

    If $f'(a) = 0$: If $f''(a) > 0$, it's a local minimum. If $f''(a) < 0$, it's a local maximum. If $f''(a) = 0$, the test is inconclusive.

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    What should you do if the second derivative test is inconclusive ($f''(x)=0$)?

    You must test the sign of the first derivative, $f'(x)$, on either side of the stationary point. This will determine if it's a point of inflection, or a max/min that the second derivative test couldn't identify.

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    How is the chain rule expressed for connected rates of change?

    If two variables, $y$ and $x$, both depend on time $t$, their rates of change are related by $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$.

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    What is the derivative of $\ln(kx)$?

    The derivative is $\frac{1}{x}$. Using the chain rule, $\frac{d}{dx}(\ln(kx)) = \frac{1}{kx} \times k = \frac{1}{x}$. The constant inside the logarithm does not affect the derivative.

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    What is the derivative of $e^{ax+b}$?

    Using the chain rule, the derivative is $a e^{ax+b}$. The derivative of the power multiplies the original function.

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    Common Trap: Differentiating $y = 3x \sin x$.

    You must use the product rule. Let $u=3x$ and $v=\sin x$. $\frac{dy}{dx} = (3x)(\cos x) + (\sin x)(3) = 3x \cos x + 3 \sin x$. A common error is to differentiate each part separately.

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    What is the difference between an increasing function and a strictly increasing function?

    A function $f(x)$ is increasing on an interval if $f'(x) \ge 0$ for all $x$ in the interval. It is strictly increasing if $f'(x) > 0$. The distinction allows for stationary points of inflection within an increasing interval.