1.2 Ableitungsregeln
Aus Online Mathematik Brückenkurs 2
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:
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(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
ddx(x2ex)=2xex+x2ex=(2x+x2)ex .ddx(xsinx)=1 .sinx+xcosx=sinx+xcosx
ddx(xlnx−x)=1 .lnx+xx1−1=lnx+1−1=lnx
ddxtanx=ddxsinxcosx=(cosx)2cosxcosx−sinx(−sinx)
=cos2xcos2x+sin2x=1cos2x .ddx x1+x=(
x)21
x−(1+x)12
x=x2x2
x−12
x−x2
x
=x2 .xx−1=x−12x
x
ddxxex1+x=(1+x)2(1 ex+xex)(1+x)−xex
1
=(1+x)2ex+xex+xex+x2ex−xex=(1+x)2(1+x+x2)ex .
Derivatives of composite functions
A function g(x)
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This rule is commonly called the chain rule and may, depending on the notation, be written in different ways. In the above example, if we put
One usually says that the composite function y consists of the outer function f and the inner function g. Analogously
Example 2
In the function
is an outer function, and | an inner function, | ||
is an outer derivative and | an inner derivative. |
The derivative of the function y with respect to x is given by the chain rule, and is
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
Do not forget to use the product and quotient rules where necessary.
Example 3
f(x)=sin(3x2+1)
Outer derivative: Inner derivative:cos(3x2+1)6x
f (x)=cos(3x2+1)
6x=6xcos(3x2+1)
y=5ex2
Outer derivative: Inner derivative:5ex22x
y =5ex2
2x=10xex2
f(x)=exsinx
Outer derivative: Inner derivative:exsinx1 sinx+xcosx
f (x)=exsinx(sinx+xcosx)
s(t)=t2cos(lnt)
s (t)=2tcos(lnt)+t2
−sin(lnt)t1
=2tcos(lnt)−tsin(lnt)
ddxax=ddx elna
x=ddxelna
x=elna
xlna=axlna
ddxxa=ddx elnx
a=ddxealnx=ealnx
ax1=xa
ax−1=axa−1
The chain rule also can be used repeatedly on a function that is composed at several levels. For example, the function g(h(x))
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Example 4
ddxsin32x=ddx(sin2x)3=3(sin2x)2ddxsin2x=3(sin2x)2cos2xddx(2x)
=3sin22xcos2x 2=6sin22xcos2x
ddxsin (x2−3x)4
=cos
(x2−3x)4
ddx(x2−3x)4
=cos (x2−3x)4
4(x2−3x)3ddx(x2−3x)
=cos (x2−3x)4
4(x2−3x)3(2x−3)
ddxsin4(x2−3x)=ddx sin(x2−3x)
4
=4sin3(x2−3x)ddxsin(x2−3x)
=4sin3(x2−3x)cos(x2−3x)ddx(x2−3x)
\displaystyle \phantom{\frac{d}{dx}\,\sin^4 (x^2 -3x)}{} = 4 \sin^3 (x^2 - 3x) \,\cos (x^2 -3x)\, (2x-3)- \displaystyle \frac{d}{dx}\,\Bigl ( e^{\sqrt{x^3-1}}\,\Bigr)
= e^{\sqrt{x^3-1}} \, \frac{d}{dx}\,\sqrt{x^3-1}
= e^{\sqrt{x^3-1}} \, \frac{1}{2 \sqrt{x^3-1}}
\, \frac{d}{dx}\,(x^3-1)
\vphantom{\Biggl(}
\displaystyle \phantom{\displaystyle \frac{d}{dx}\,\Bigl ( e^{\sqrt{x^3-1}}\,\Bigr)}{} = e^{\sqrt{x^3-1}} \, \frac{1}{2 \sqrt{x^3-1}} \times 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} \, \frac{d}{dx}\,(x^2-1) = 3\,e^{x^2 -1} \times 2x = 6x\,e^{x^2 -1}\vphantom{\biggl(}
\displaystyle f^{\,\prime\prime}(x) = 6\,e^{x^2 -1} + 6x\,e^{x^2 -1} \times 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 \frac{d}{dx}\,( e^x \sin x) = e^x \sin x + e^x \cos x
= e^x (\sin x + \cos x)
\vphantom{\Bigl(}
\displaystyle \frac{d^2}{dx^2}(e^x\sin x) = \frac{d}{dx}\,\bigl(e^x (\sin x + \cos x)\bigr) \vphantom{\Bigl(} \displaystyle \phantom{\frac{d^2}{dx^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 \frac{d^3}{dx^3} ( e^x \sin x) = \frac{d}{dx}\,(2\,e^x \cos x) \vphantom{\Bigl(} \displaystyle \phantom{\frac{d^3}{dx^3} ( e^x \sin x)}{} = 2\,e^x \cos x + 2\,e^x (-\sin x) = 2\,e^x ( \cos x - \sin x )