1.2 Ableitungsregeln
Aus Online Mathematik Brückenkurs 2
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<div class="regel"> | <div class="regel"> | ||
'''Rules of differentiation for products and quotients: ''' | '''Rules of differentiation for products and quotients: ''' | ||
| - | {{ | + | {{Displayed math||<math>\begin{align*} D\,\bigl(\,f(x) \cdot g(x) \bigr) &= f^{\,\prime}(x) \cdot g(x) + f(x) \cdot g'(x)\\[4pt] D\,\Bigl( \frac{f(x)}{g(x)} \Bigr) &= \frac{f^{\,\prime}(x)\cdot g(x) - f(x)\cdot g'(x)}{\bigl(g(x)\bigr)^2} \end{align*}</math>}} |
</div> | </div> | ||
(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!) | (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!) | ||
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A function <math>y=f(g)</math> where the variable ''g'' in turn, is dependent on a variable ''x'' takes the form <math>y=f \bigl( g(x)\bigr)</math> and is called a composite function. If one differentiates a composite function with respect to the independent variable ''x'', one uses the following rule | A function <math>y=f(g)</math> where the variable ''g'' in turn, is dependent on a variable ''x'' takes the form <math>y=f \bigl( g(x)\bigr)</math> and is called a composite function. If one differentiates a composite function with respect to the independent variable ''x'', one uses the following rule | ||
| - | {{ | + | {{Displayed math||<math>y'(x) = f^{\,\prime}\bigl( g(x) \bigr) |
\cdot g'(x)\,\mbox{.}</math>}} | \cdot g'(x)\,\mbox{.}</math>}} | ||
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 <math>y=f(u)</math> and <math>u=g(x)</math> the chain rule can be written | 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 <math>y=f(u)</math> and <math>u=g(x)</math> the chain rule can be written | ||
| - | {{ | + | {{Displayed math||<math>\frac{dy}{dx} |
= \frac{dy}{du} \cdot \frac{du}{dx}\,\mbox{.}</math>}} | = \frac{dy}{du} \cdot \frac{du}{dx}\,\mbox{.}</math>}} | ||
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The derivative of the function ''y'' with respect to ''x'' is given by the chain rule, and is | The derivative of the function ''y'' with respect to ''x'' is given by the chain rule, and is | ||
| - | {{ | + | {{Displayed math||<math>\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} |
= 4 u^3 \cdot (2x +2) = 4(x^2 + 2x)^3 \cdot (2x +2)\,\mbox{.}</math>}} | = 4 u^3 \cdot (2x +2) = 4(x^2 + 2x)^3 \cdot (2x +2)\,\mbox{.}</math>}} | ||
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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 | 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 | ||
| - | {{ | + | {{Displayed math||<math>(\text{outer derivative}) |
\cdot (\text{ inner derivative})\,\mbox{.}</math>}} | \cdot (\text{ inner derivative})\,\mbox{.}</math>}} | ||
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| - | {{ | + | {{Displayed math||<math>y'= f^{\,\prime} \bigl ( g(h(x))\bigr) |
\cdot g'(h(x)) \cdot h'(x)\,\mbox{.}</math>}} | \cdot g'(h(x)) \cdot h'(x)\,\mbox{.}</math>}} | ||
Version vom 08:24, 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:
| \displaystyle \begin{align*} D\,\bigl(\,f(x) \cdot g(x) \bigr) &= f^{\,\prime}(x) \cdot g(x) + f(x) \cdot g'(x)\\[4pt] D\,\Bigl( \frac{f(x)}{g(x)} \Bigr) &= \frac{f^{\,\prime}(x)\cdot g(x) - f(x)\cdot g'(x)}{\bigl(g(x)\bigr)^2} \end{align*} |
(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
- \displaystyle D\,(x^2 e^x) = 2x\cdot e^x + x^2\cdot e^x = (2x +x^2)\,e^x\,.
- \displaystyle D\,(x \sin x) = 1\cdot \sin x + x\cdot\cos x = \sin x + x \cos x\,.
- \displaystyle D\,(x \ln x -x) = 1 \cdot \ln x + x\cdot \frac{1}{x} - 1 = \ln x + 1 -1 = \ln x\,.
- \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}\,. - \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}}\,. - \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
\displaystyle y'(x) = f^{\,\prime}\bigl( g(x) \bigr)
\cdot g'(x)\,\mbox{.}
|
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
\displaystyle \frac{dy}{dx}
= \frac{dy}{du} \cdot \frac{du}{dx}\,\mbox{.}
|
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
\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
= 4 u^3 \cdot (2x +2) = 4(x^2 + 2x)^3 \cdot (2x +2)\,\mbox{.}
|
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
\displaystyle (\text{outer derivative})
\cdot (\text{ inner derivative})\,\mbox{.}
|
Do not forget to use the product and quotient rules where necessary.
Example 3
- \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) - \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} - \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) - \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) - \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
- \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
\displaystyle y'= f^{\,\prime} \bigl ( g(h(x))\bigr)
\cdot g'(h(x)) \cdot h'(x)\,\mbox{.}
|
Example 4
- \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 - \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) - \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) - \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
- \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) - \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 - \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 )
