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9709 · 1.7

Differentiation flashcards

Revision flashcards for Cambridge 9709 Differentiation (syllabus 1.7). Flip, recall, then mark a real past-paper question.

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    What is differentiation?

    It is the process of finding the derivative, or gradient function, of a function. This new function tells you the instantaneous rate of change or the gradient of the original function at any point.

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    What are the two main notations for the derivative of $y = f(x)$?

    Leibniz's notation: $\frac{dy}{dx}$ (read as 'dee-why by dee-ex'). Lagrange's notation: $f'(x)$ (read as 'f-dashed-x' or 'f-prime-x').

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    What is the Power Rule for differentiation?

    For a function of the form $y = ax^n$, the derivative is $\frac{dy}{dx} = anx^{n-1}$. You multiply by the power and then reduce the power by one.

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    What is the derivative of a constant term, e.g., $y = 7$?

    The derivative of any constant is zero. A line $y=c$ is horizontal, so its gradient is always 0.

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    How do you differentiate a function with multiple terms, like $y = x^3 + 4x^2$?

    Differentiate each term separately and add or subtract them as in the original function. For $y = x^3 + 4x^2$, $\frac{dy}{dx} = 3x^2 + 8x$.

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    What is a tangent to a curve?

    A straight line that 'just touches' the curve at a single point. The gradient of the tangent is equal to the gradient of the curve at that point.

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    What is a normal to a curve?

    A straight line that is perpendicular to the tangent at the point of contact. Its gradient is the negative reciprocal of the tangent's gradient: $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$.

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    What is a stationary point (or turning point)?

    A point on the curve where the gradient is zero. At this point, the tangent is horizontal. This occurs at local maxima, local minima, and points of inflection.

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    How do you find the coordinates of a stationary point?

    1. Differentiate the function to find $\frac{dy}{dx}$. 2. Set $\frac{dy}{dx} = 0$ and solve for $x$. 3. Substitute the $x$-value(s) back into the original equation for the curve to find the corresponding $y$-coordinate(s).

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    How do you prepare an expression like $y = \frac{3}{x^2}$ for differentiation?

    Rewrite it using negative indices before applying the power rule. $y = 3x^{-2}$. Then differentiate to get $\frac{dy}{dx} = 3(-2)x^{-2-1} = -6x^{-3} = -\frac{6}{x^3}$.

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    How do you prepare an expression like $y = 5\sqrt{x}$ for differentiation?

    Rewrite it using fractional indices before applying the power rule. $y = 5x^{1/2}$. Then differentiate to get $\frac{dy}{dx} = 5(\frac{1}{2})x^{1/2-1} = \frac{5}{2}x^{-1/2} = \frac{5}{2\sqrt{x}}$.