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General information
Types of DEs
Manipulation
  • Introduction
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  • Linearity
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  • Product rule
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  • Quotent rule
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  • Chain rule
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  • Separation of variables

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    Examples

    Quotient rule

    In calculus, the quotient rule is a method of finding the derivative of a function which is the quotient of two other functions for which derivatives exist.

    If the function one wishes to differentiate, f(x), can be written as

    f(x) = \frac{g(x)}{h(x)}

    and h(x) ≠ 0; then, the rule states that the derivative of g(x) / h(x) is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator:

    f'(x)=\frac{g'(x)h(x) - g(x)h'(x)}{{h(x)}^2}.

    Or more precisely; for all x in some open set containing the number a, with h(a) ≠ 0; and, such that g '(a) and h '(a) both exist; then, f '(a) exists as well:

    f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{h(a)^2}

    Examples

    The derivative of (4x - 2) / (x2 + 1) = [(x2 + 1)(4) - (4x - 2)(2x)] / (x2 + 1)2 = [(4x2 + 4) - (8x2 - 4x)] / (x2 + 1)2 = [-4x2 + 4x + 4] / (x2 + 1)2

    The derivative of [sin(x)] / x2 (when x ≠ 0) is ([cos(x)]x2 - [sin(x)](2x)) / x4.

    For more information regarding the derivatives of trigonometric functions, see: derivative.

    Another example is:

    f(x) = {2x^2 \over x^3}

    whereas g(x) = 2x2 and h(x) = x3, and g′(x) = 4x and h′(x) = 3x2.

    The derivative of f(x) is determined as follows:

    f'(x) = {\left[\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \right)\right] \over \left(x^3\right)^2}
    f'(x) = {4x^4 - 6x^4 \over x^6}
    f'(x) = {-2x^4 \over x^6}
    f'(x) = {-2 \over x^2}

    Proof

    A proof of this rule can be derived from Newton's difference quotient:

    \mbox{let }f(x) = \frac{g(x)}{h(x)}
    where h(x) ≠ 0 and g and h are differentiable.
    f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\frac{g(x + \Delta x)}{h(x + \Delta x)} - \frac{g(x)}{h(x)}}{\Delta x}
    = \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left[ \frac{g(x+\Delta x)h(x)-g(x)h(x+\Delta x)}{h(x)h(x+\Delta x)} \right]
    = \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left[ \frac{[g(x+\Delta x)h(x)-g(x)h(x)]-[g(x)h(x+\Delta x)-g(x)h(x)]}{h(x)h(x+\Delta x)} \right]
    = \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left[ \frac{h(x)[g(x+\Delta x)-g(x)]-g(x)[h(x+\Delta x)-h(x)]}{h(x)h(x+\Delta x)} \right]
    = \lim_{\Delta x \to 0} \frac{\frac{g(x+\Delta x)-g(x)}{\Delta x}h(x)-g(x)\frac{h(x+\Delta x)-h(x)}{\Delta x}}{h(x)h(x+\Delta x)}
    = \frac{\lim_{\Delta x \to 0} \left(\frac{g(x+\Delta x)-g(x)}{\Delta x}\right)h(x) - g(x) \lim_{\Delta x \to 0} \left(\frac{h(x+\Delta x)-h(x)}{\Delta x}\right)}{h(x)h(\lim_{\Delta x \to 0} (x+\Delta x))}
    = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}

    Alternate informal proof

    Using only the product rule:

    f(x)=\frac{g(x)}{h(x)}
    g(x) = f(x)h(x)
    g'(x) = f'(x)h(x) + f(x)h'(x)

    The rest is simple algebra to make f'(x) the only term on the left hand side of the equation and to remove f(x) from the right side of the equation.

    f'(x)=\frac{g'(x) - f(x)h'(x)}{h(x)} = \frac{g'(x) - \frac{g(x)}{h(x)}\cdot h'(x)}{h(x)}
    f'(x)=\frac{g'(x)h(x) - g(x)h'(x)}{\left(h(x)\right)^2}

    Mnemonic

    It is often memorized as a rhyme type song. "Lo-dee-hi minus hi-dee-lo all over lo-lo"; Lo being the denominator, Hi being the numerator and D being the derivative.

    Solving linear equations

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