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Version vom 08:20, 17. Sep. 2008

       Theory          Exercises      

Contents:

  • Derivative of a product and quotient
  • Derivative of a composite function (chain rule)
  • Higher order derivatives

Learning outcomes:

After this section, you will have learned :

  • To be able, in principle, to differentiate any function composed of elementary functions

Differentiation of products and quotients

Using the definition of a derivative one can obtain the rules for differentiation of products and quotients of functional expressions:

Rules of differentiation for products and quotients:

  1. REDIRECT Template:Abgesetzte Formel

(Note that the derivatives of products and quotients are not as simple as the derivatives of sums and differences, where one can differentiate the functional expression term by term, i.e. individually!)

Example 1

  1. \displaystyle D\,(x^2 e^x) = 2x\cdot e^x + x^2\cdot e^x = (2x +x^2)\,e^x\,.
  2. \displaystyle D\,(x \sin x) = 1\cdot \sin x + x\cdot\cos x = \sin x + x \cos x\,.
  3. \displaystyle D\,(x \ln x -x) = 1 \cdot \ln x + x\cdot \frac{1}{x} - 1 = \ln x + 1 -1 = \ln x\,.
  4. \displaystyle D\,\tan x = D\,\frac{\sin x}{\cos x} = \frac{ \cos x \cdot \cos x - \sin x \cdot (-\sin x)}{(\cos x)^2} \vphantom{\biggl(}
    \displaystyle \phantom{D\,\tan x}{} = \frac{\cos^2 x + \sin^2 x }{ \cos^2 x} = \frac{1}{\cos^2 x}\,.
  5. \displaystyle D\,\frac{1+x}{\sqrt{x}} = \frac{\displaystyle 1 \cdot \sqrt{x} - (1+x) \cdot \frac{1}{2\sqrt{x}}}{(\sqrt{x}\,)^2} = \frac{\displaystyle\frac{2x}{2\sqrt{x}} - \frac{1}{2\sqrt{x}} - \frac{x}{2\sqrt{x}}}{x} \vphantom{\biggl(}
    \displaystyle \phantom{D\,\frac{1+x}{\sqrt{x}}}{} = \frac {\displaystyle \frac {x-1}{2\sqrt{x}}}{x} = \frac{x-1}{2x\sqrt{x}}\,.
  6. \displaystyle D\,\frac{x\,e^x}{1+x} = \frac{(1\cdot e^x + x\cdot e^x)(1+x) - x\,e^x \cdot 1}{(1+x)^2} \vphantom{\Biggl(}
    \displaystyle \phantom{D\,\frac{x\,e^x}{1+x}}{} = \frac{ e^x + x\,e^x + x\,e^x + x^2\,e^x - x\,e^x}{(1+x)^2} = \frac{(1 + x + x^2)\,e^x} {(1+x)^2}\,.


Derivatives of composite functions

A function \displaystyle y=f(g) where the variable g in turn, is dependent on a variable x takes the form \displaystyle y=f \bigl( g(x)\bigr) and is called a composite function. If one differentiates a composite function with respect to the independent variable x, one uses the following rule

  1. REDIRECT Template:Abgesetzte Formel

This rule is commonly called the chain rule and may, depending on the notation be written in different ways. If we in the above example put \displaystyle y=f(u) and \displaystyle u=g(x) the chain rule can be written

  1. REDIRECT Template:Abgesetzte Formel

One usually says that the composite function y consists of the outer function f and the inner function g. Analogously \displaystyle f^{\,\prime} is said to be the outer derivative and \displaystyle g' the inner derivative.


Example 2

In the function \displaystyle y=(x^2 + 2x)^4

\displaystyle y=u^4 is an outer function, and \displaystyle u=x^2+2x an inner function,
\displaystyle \dfrac{dy}{du}=4u^3 is an outer derivative and \displaystyle \dfrac{du}{dx}=2x+2 an inner derivative.

The derivative of the function y with respect to x is given by the chain rule, and is

  1. REDIRECT Template:Abgesetzte Formel

When one has become accustomed to using the chain rule one seldom introduces new symbols for the inner and outer functions, but one learns to recognise them intuitively and can differentiate ”immediately”, according to the rule

  1. REDIRECT Template:Abgesetzte Formel

Do not forget to use the product and quotient rules where necessary.

Example 3

  1. \displaystyle f(x) = \sin (3x^2 + 1)

    \displaystyle \begin{array}{ll} \text{Outer derivative:} & \cos (3x^2 +1)\\ \text{ Inner derivative:} & 6x \end{array}

    \displaystyle f^{\,\prime}(x) = \cos (3x^2 + 1) \cdot 6x = 6x \cos (3x^2 +1)
  2. \displaystyle y = 5 \, e^{x^2}

    \displaystyle \begin{array}{ll} \text{Outer derivative:} & 5\,e^{x^2}\\ \text{ Inner derivative:} & 2x \end{array}

    \displaystyle y' = 5 \, e^{x^2} \cdot 2x = 10x\, e^{x^2}
  3. \displaystyle f(x) = e^{x\cdot \sin x}

    \displaystyle \begin{array}{ll} \text{Outer derivative:} & e^{x\cdot \sin x}\\ \text{ Inner derivative:} & 1\cdot \sin x + x \cos x \end{array}

    \displaystyle f^{\,\prime}(x) = e^{x\cdot \sin x} (\sin x + x \cos x)
  4. \displaystyle s(t) = t^2 \cos (\ln t)

    \displaystyle s'(t) = 2t \cdot \cos (\ln t) + t^2 \cdot\Bigl(-\sin (\ln t) \cdot\frac{1}{t}\Bigr) = 2t \cos (\ln t) - t \sin (\ln t)
  5. \displaystyle D\,a^x = D\,\bigl( e^{\ln a} \bigr)^x = D\,e^{\ln a \cdot x} = e^{\ln a \cdot x} \cdot \ln a = a^x \cdot \ln a
  6. \displaystyle D\,x^a = D\,\bigl( e^{\ln x} \bigr)^a = D\,e^{ a \cdot \ln x } = e^{a \cdot \ln x} \cdot a \cdot \frac{1}{x} = x^a \cdot a \cdot x^{-1} = ax^{a-1}

The chain rule also can be used repeatedly on a function that is composed at several levels. For example, the function \displaystyle y= f \bigl( g(h(x))\bigr) has the derivative


  1. REDIRECT Template:Abgesetzte Formel


Example 4

  1. \displaystyle D\,\sin^3 2x = D\,(\sin 2x)^3 = 3(\sin 2x)^2 \cdot D\,\sin 2x = 3(\sin 2x)^2 \cdot \cos 2x \cdot D\,(2x) \vphantom{\Bigl(}
    \displaystyle \phantom{D\,\sin^3 2x}{}= 3 \sin^2 2x\cdot\cos 2x\cdot 2 = 6 \sin^2 2x\,\cos 2x
  2. \displaystyle D\,\sin \bigl((x^2 -3x)^4 \bigr) = \cos \bigl((x^2 -3x)^4\bigr) \cdot D\,(x^2 -3x)^4 \vphantom{\Bigl(}
    \displaystyle \phantom{D\,\sin \bigl((x^2 -3x)^4 \bigr)}{} = \cos \bigl((x^2 -3x)^4\bigr)\cdot 4 (x^2 -3x)^3 \cdot D\,(x^2-3x) \vphantom{\Bigl(}
    \displaystyle \phantom{D\,\sin \bigl((x^2 -3x)^4 \bigr)}{} = \cos \bigl((x^2 -3x)^4\bigr)\cdot 4 (x^2 -3x)^3 \cdot (2x-3)
  3. \displaystyle D\,\sin^4 (x^2 -3x) = D\,\bigl( \sin (x^2 -3x) \bigr)^4 \vphantom{\Bigl(}
    \displaystyle \phantom{D\,\sin^4 (x^2 -3x)}{} = 4 \sin^3 (x^2 - 3x) \cdot D\,\sin(x^2-3x) \vphantom{\Bigl(}
    \displaystyle \phantom{D\,\sin^4 (x^2 -3x)}{} = 4 \sin^3 (x^2 - 3x) \cdot\cos (x^2 -3x) \cdot D(x^2 -3x) \vphantom{\Bigl(}
    \displaystyle \phantom{D\,\sin^4 (x^2 -3x)}{} = 4 \sin^3 (x^2 - 3x) \cdot\cos (x^2 -3x)\cdot (2x-3)
  4. \displaystyle D\,\Bigl ( e^{\sqrt{x^3-1}}\,\Bigr) = e^{\sqrt{x^3-1}} \cdot D\,\sqrt{x^3-1} = e^{\sqrt{x^3-1}} \cdot \frac{1}{2 \sqrt{x^3-1}} \cdot D\,(x^3-1) \vphantom{\Biggl(}
    \displaystyle \phantom{\displaystyle D\,\Bigl ( e^{\sqrt{x^3-1}}\,\Bigr)}{} = e^{\sqrt{x^3-1}} \cdot \frac{1}{2 \sqrt{x^3-1}} \cdot 3 x^2 = \frac { 3 x^2 e^{\sqrt{x^3-1}}} {2 \sqrt{x^3-1}} \vphantom{\dfrac{\dfrac{()^2}{()}}{()}}


Higher order derivatives

If a function is differentiable more than once, one can consider higher derivatives like the second derivative, third derivative, and so on.

The second derivative usually is written as \displaystyle f^{\,\prime\prime} (sometimes referred to as "double-prime"), while the third, fourth, etc. derivatives, are written as \displaystyle f^{\,(3)}, \displaystyle f^{\,(4)} and so on.

Other usual notations for these quantities are \displaystyle D^2 f, \displaystyle D^3 f, \displaystyle \ldots\,, \displaystyle \frac{d^2 y}{dx^2}, \displaystyle \frac{d^3 y}{dx^3}, \displaystyle \ldots.

Example 5

  1. \displaystyle f(x) = 3\,e^{x^2 -1}
    \displaystyle f^{\,\prime}(x) = 3\,e^{x^2 -1} \cdot D\,(x^2-1) = 3\,e^{x^2 -1} \cdot 2x = 6x\,e^{x^2 -1}\vphantom{\biggl(}
    \displaystyle f^{\,\prime\prime}(x) = 6\,e^{x^2 -1} + 6x\,e^{x^2 -1} \cdot 2x = 6\,e^{x^2 -1}\,(1+ 2x^2)
  2. \displaystyle y = \sin x\,\cos x
    \displaystyle \frac{dy}{dx} = \cos x\,\cos x + \sin x\,(- \sin x) = \cos^2 x - \sin^2 x\vphantom{\Biggl(}
    \displaystyle \frac{d^2 y}{dx^2} = 2 \cos x\,(-\sin x) - 2 \sin x \cos x = -4 \sin x \cos x
  3. \displaystyle D\,( e^x \sin x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x) \vphantom{\Bigl(}
    \displaystyle D^2(e^x\sin x) = D\,\bigl(e^x (\sin x + \cos x)\bigr) \vphantom{\Bigl(} \displaystyle \phantom{D^2(e^x\sin x)}{} = e^x (\sin x + \cos x) + e^x (\cos x - \sin x) = 2\,e^x \cos x \vphantom{\biggl(}
    \displaystyle D^3 ( e^x \sin x) = D\,(2\,e^x \cos x) \vphantom{\Bigl(} \displaystyle \phantom{D^3 ( e^x \sin x)}{} = 2\,e^x \cos x + 2\,e^x (-\sin x) = 2\,e^x ( \cos x - \sin x )