1.2 Rules of differentiation

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__NOTOC__
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{{Selected tab|[[1.2 Rules of differentiation|Theory]]}}
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{{Not selected tab|[[1.2 Exercises|Exercises]]}}
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Contents:'''
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* Derivata av en produkt och kvot
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* Derivative of a product and quotient
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* Derivata av en sammansatt funktion (kedjeregeln)
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* Derivative of a composite function (chain rule)
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* Högre ordningars derivata
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* Higher order derivatives
}}
}}
{{Info|
{{Info|
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'''Färdigheter:'''
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'''Learning outcomes:'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned :
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* I princip kunna derivera vilken elementär funktion som helst.
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* To be able, in principle, to differentiate any function composed of elementary functions
}}
}}
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== Derivering av produkt och kvot ==
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== Differentiation of products and quotients==
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Med hjälp av derivatans definition kan man också härleda deriveringsregler för produkter och kvoter av funktionsuttryck:
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Using the definition of a derivative one can obtain the rules for differentiation of products and quotients of functional expressions:
<div class="regel">
<div class="regel">
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'''Deriveringsregler för produkter och kvoter:'''
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'''Rules of differentiation for products and quotients: '''
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{{Fristående formel||<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>}}
+
{{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>
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(Observera att derivering av produkter och kvoter '''inte''' är så enkelt som derivering av summor och differenser, där man kan derivera funktionsuttrycken termvis, dvs. var för sig!)
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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!)
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
<ol type="a">
<ol type="a">
<li><math>D\,(x^2 e^x) = 2x\cdot e^x + x^2\cdot e^x
<li><math>D\,(x^2 e^x) = 2x\cdot e^x + x^2\cdot e^x
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== Derivering av sammansatta funktioner ==
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== Derivatives of composite functions ==
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En funktion <math>y=f(g)</math> där variabeln ''g'' i sin tur är beroende av en variabel ''x'' får formen <math>y=f \bigl( g(x)\bigr)</math> och kallas sammansatt funktion. Om man deriverar en sammansatt funktion med avseende på den oberoende variabeln ''x'', använder man följande regel:
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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
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{{Fristående formel||<math>y'(x) = f^{\,\prime}\bigl( g(x) \bigr)
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{{Displayed math||<math>y'(x) = f^{\,\prime}\bigl( g(x) \bigr)
\cdot g'(x)\,\mbox{.}</math>}}
\cdot g'(x)\,\mbox{.}</math>}}
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Denna regel brukar kallas kedjeregeln och kan beroende på val av symboler skrivas på olika sätt. Om vi i ovanstående t.ex. sätter <math>y=f(u)</math> och <math>u=g(x)</math> kan kedjeregeln skrivas
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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
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{{Fristående formel||<math>\frac{dy}{dx}
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{{Displayed math||<math>\frac{dy}{dx}
= \frac{dy}{du} \cdot \frac{du}{dx}\,\mbox{.}</math>}}
= \frac{dy}{du} \cdot \frac{du}{dx}\,\mbox{.}</math>}}
-
Man brukar säga att den sammansatta funktionen ''y'' består av den ''yttre'' funktionen ''f'' och den ''inre'' funktionen ''g''. Analogt kallas <math>f^{\,\prime}</math> för den ''yttre derivatan'' och <math>g'</math> den ''inre derivatan''.
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One usually says that the composite function ''y'' consists of the ''outer'' function ''f'' and the ''inner'' function ''g''. Analogously <math>f^{\,\prime}</math> is said to be the ''outer derivative'' and <math>g'</math> the ''inner derivative''.
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
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För funktionen <math>y=(x^2 + 2x)^4</math> är
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In the function <math>y=(x^2 + 2x)^4</math>
<center>
<center>
{| align="center" cellspacing="3em" cellpadding="0"
{| align="center" cellspacing="3em" cellpadding="0"
| align="left" |<math>y=u^4</math>
| align="left" |<math>y=u^4</math>
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| align="left" |yttre funktionen, och
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| align="left" | is an outer function, and
| align="left" |<math>u=x^2+2x</math>
| align="left" |<math>u=x^2+2x</math>
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| align="left" |inre funktionen.
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| align="left" |an inner function,
|-
|-
| align="left" |<math>\dfrac{dy}{du}=4u^3</math>
| align="left" |<math>\dfrac{dy}{du}=4u^3</math>
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| align="left" |yttre derivata, och
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| align="left" | is an outer derivative and
| align="left" |<math>\dfrac{du}{dx}=2x+2</math>
| align="left" |<math>\dfrac{du}{dx}=2x+2</math>
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| align="left" |inre derivata
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| align="left" |an inner derivative.
|}
|}
</center>
</center>
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Derivatan av funktionen ''y'' med avseende på ''x'' blir enligt kedjeregeln
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The derivative of the function ''y'' with respect to ''x'' is given by the chain rule, and is
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{{Fristående formel||<math>\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
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{{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>}}
</div>
</div>
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När man vant sig vid kedjeregeln inför man sällan nya beteckningar för yttre och inre funktion, utan man lär sig känna igen dessa och deriverar ”rakt på”, enligt mönstret
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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
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{{Fristående formel||<math>(\text{yttre derivata})
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{{Displayed math||<math>(\text{outer derivative})
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\cdot (\text{inre derivata})\,\mbox{.}</math>}}
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\cdot (\text{ inner derivative})\,\mbox{.}</math>}}
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Kom ihåg att även använda produkt- eller kvotregeln när detta är nödvändigt.
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Do not forget to use the product and quotient rules where necessary.
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
<ol type="a">
<ol type="a">
<li><math> f(x) = \sin (3x^2 + 1)</math><br><br>
<li><math> f(x) = \sin (3x^2 + 1)</math><br><br>
<math>\begin{array}{ll}
<math>\begin{array}{ll}
-
\text{Yttre derivatan:} & \cos (3x^2 +1)\\
+
\text{Outer derivative:} & \cos (3x^2 +1)\\
-
\text{Inre derivatan:} & 6x
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\text{ Inner derivative:} & 6x
\end{array}</math><br><br>
\end{array}</math><br><br>
<math>f^{\,\prime}(x) = \cos (3x^2 + 1) \cdot 6x
<math>f^{\,\prime}(x) = \cos (3x^2 + 1) \cdot 6x
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<li><math> y = 5 \, e^{x^2}</math><br><br>
<li><math> y = 5 \, e^{x^2}</math><br><br>
<math>\begin{array}{ll}
<math>\begin{array}{ll}
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\text{Yttre derivatan:} & 5\,e^{x^2}\\
+
\text{Outer derivative:} & 5\,e^{x^2}\\
-
\text{Inre derivatan:} & 2x
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\text{ Inner derivative:} & 2x
\end{array}</math><br><br>
\end{array}</math><br><br>
<math>y' = 5 \, e^{x^2} \cdot 2x = 10x\, e^{x^2}</math>
<math>y' = 5 \, e^{x^2} \cdot 2x = 10x\, e^{x^2}</math>
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<li><math> f(x) = e^{x\cdot \sin x}</math><br><br>
<li><math> f(x) = e^{x\cdot \sin x}</math><br><br>
<math>\begin{array}{ll}
<math>\begin{array}{ll}
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\text{Yttre derivatan:} & e^{x\cdot \sin x}\\
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\text{Outer derivative:} & e^{x\cdot \sin x}\\
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\text{Inre derivatan:} & 1\cdot \sin x + x \cos x
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\text{ Inner derivative:} & 1\cdot \sin x + x \cos x
\end{array}</math><br><br>
\end{array}</math><br><br>
<math>f^{\,\prime}(x) = e^{x\cdot \sin x} (\sin x + x \cos x)</math>
<math>f^{\,\prime}(x) = e^{x\cdot \sin x} (\sin x + x \cos x)</math>
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</div>
</div>
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Kedjeregeln kan även användas upprepade gånger på en funktion som är sammansatt i flera steg. Exempelvis funktionen <math>y= f \bigl( g(h(x))\bigr)</math> har derivatan
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The chain rule also can be used repeatedly on a function that is composed at several levels. For example, the function <math>y= f \bigl( g(h(x))\bigr)</math> has the derivative
 +
-
{{Fristående formel||<math>y'= f^{\,\prime} \bigl ( g(h(x))\bigr)
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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>}}
<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
<ol type="a">
<ol type="a">
<li><math> D\,\sin^3 2x = D\,(\sin 2x)^3
<li><math> D\,\sin^3 2x = D\,(\sin 2x)^3
Line 172: Line 180:
<li><math> D\,\sin \bigl((x^2 -3x)^4 \bigr)
<li><math> D\,\sin \bigl((x^2 -3x)^4 \bigr)
= \cos \bigl((x^2 -3x)^4\bigr)
= \cos \bigl((x^2 -3x)^4\bigr)
-
\cdot D\,\bigl((x^2 -3x)^4\bigr)
+
\cdot D\,(x^2 -3x)^4
\vphantom{\Bigl(}</math><br>
\vphantom{\Bigl(}</math><br>
<math>\phantom{D\,\sin \bigl((x^2 -3x)^4 \bigr)}{}
<math>\phantom{D\,\sin \bigl((x^2 -3x)^4 \bigr)}{}
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== Derivator av högre ordningar ==
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==Higher order derivatives ==
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Om en funktion är deriverbar mer än en gång så pratar man om funktionens andra-, tredjederivata, osv.
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If a function is differentiable more than once, one can consider higher derivatives like the second derivative, third derivative, and so on.
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Andraderivatan brukar betecknas <math>f^{\,\prime\prime}</math> (läses "f-biss"), medan tredje-, fjärdederivatan, etc, betecknas <math>f^{\,(3)}</math>, <math>f^{\,(4)}</math> osv.
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The second derivative usually is written as <math>f^{\,\prime\prime}</math> (sometimes referred to as "double-prime"), while the third, fourth, etc. derivatives, are written as <math>f^{\,(3)}</math>, <math>f^{\,(4)}</math> and so on.
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Även beteckningarna <math>D^2 f</math>, <math>D^3 f</math>, <math>\ldots</math> och <math>\frac{d^2 y}{dx^2}</math>, <math>\frac{d^3 y}{dx^3}</math>, <math>\ldots</math> är vanliga.
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Other usual notations for these quantities are <math>D^2 f</math>, <math>D^3 f</math>, <math>\ldots\,</math>, <math>\frac{d^2 y}{dx^2}</math>, <math>\frac{d^3 y}{dx^3}</math>, <math>\ldots</math>.
<div class="exempel">
<div class="exempel">
-
'''Exempel 5'''
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''' Example 5'''
<ol type="a">
<ol type="a">
<li><math>f(x) = 3\,e^{x^2 -1}</math> <br>
<li><math>f(x) = 3\,e^{x^2 -1}</math> <br>

Current revision

       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

  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

\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

  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


\displaystyle y'= f^{\,\prime} \bigl ( g(h(x))\bigr) 
 \cdot g'(h(x)) \cdot h'(x)\,\mbox{.}


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 )
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