IT204 - Deriatives
Definitions and Properties
Dr Robert Batzinger
Instructor Emeritus
Payap University
Chiang Mai, Thailand
15-Aug-2022
Agenda
- Definitions
- The derivative of a function
- Basic differentiation rules and rates of change
- The chain rule
- Implicit differentiation
Definitions
\[\lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}\] * Derivative of \(f(x)\) witrh respect to the variable x: (as long as the exists)
\[f^\prime(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}\] * Alternative ways of referring to derivatives:
If \(y = f(x)\):
\[y^\prime = f^\prime (x) = \frac{dy}{dx} = \frac{d}{dx} f(x) = \frac{df}{dx}\] ## Comparison
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The derivative of a function
\[y = \sqrt{x}\]
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Calculating
\[f(x) = \sqrt{x}, \quad f(x+h) = \sqrt{x + h}\] $$\[\begin{eqnarray}
\frac{f(x+h) - f(x)}{h}&=& \frac{\sqrt{x+h} - \sqrt{x}}{h}\\
&=& \frac{(x+h) -x}{h (\sqrt{x+h} +\sqrt{x})}\\
&=& \frac{1}{\sqrt{x+h} +\sqrt{x}}\\
\end{eqnarray}\]$$
\[f^\prime (x) = \lim_{x\rightarrow 0} \frac{1}{\sqrt{x+h} +\sqrt{x}} = \frac{1}{2\sqrt{x}} \]
Basic differentiation rules
Assuming \(u = f_1(x);\ v = f_2(x)\)
\(\begin{eqnarray} 1) && \hbox{Constant rule:}& f(x) = c && \frac{d}{dx} (c) = 0\\ 2) && \hbox{Power rule:}& f(x) = x^n && \frac{d}{dx} x^n = n x^{n-1}\\ 3) && \hbox{Constant multiple rule:} & f(x) = c\ u && \frac{d}{dx} c\ u = c \frac{du}{dx}\\ 4) && \hbox{Sum rule:}& \ f(x) = u + v && \frac{d}{dx} u + v = \frac{du}{dx} + \frac{dv}{dx}\\ 5) && \hbox{Product rule:}& f(x) = u\ v && \frac{d}{dx} u\ v = u \frac{dv}{dx} + v\frac{du}{dx}\\ \end{eqnarray}\)
Finding higher derivatives
\[\begin{eqnarray}
y &=& x^3 -3x^2 -x +2\\
y^\prime &=& 3x^2 - 6x-1\\
y^{\prime\prime} &=& 6x -6\\
y^{\prime\prime\prime} &=& 6\\
y^{\prime\prime\prime\prime} &=& 0\\
\end{eqnarray}\]
Finding higher derivatives
$$\[\begin{eqnarray}
y &=& \sqrt{x}= x^{\frac{1}{2}}\\
y^\prime &=& \frac{x^{-\frac{1}{2}}}{2}\\
y^{\prime\prime} &=&-\frac{x^{-\frac{3}{2}}}{4}\\
y^{\prime\prime\prime} &=& \frac{3x^{-\frac{5}{2}}}{8}\\
y^{\prime\prime\prime\prime} &=& -\frac{15x^{-\frac{7}{2}}}{16}\\
\end{eqnarray}\]$$
Exercises
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Graphic representation of higher derivatives
The chain rule
Assuming \(u = g(x)\)
\[\frac{dy}{dx} = \frac{dy}{du}\ \frac{du}{dx}\]
A Simple example
\[d\frac{d}{dx} \left(x^3 \right)^2= \frac{d}{dx} x^6= 6 x^5\] \[d\frac{d}{dx} \left(x^3 \right)^2 = (x^3)(x^3) = v\frac{du}{dx} u\frac{dv}{dx} = x^3(3x^2) + x^3(3x^2)=6x^5\] \[u = x^3;\quad\frac{d}{dx} \left(x^3 \right)^2=\frac{dy}{du} \frac{du}{dx} = \frac{(u)^2}{du} \frac{d}{dx}u= 2x^3\ 3x^2 = 6x^5\]
Differentiation
\[\begin{eqnarray}
y &=& x^\frac{1}{2} &=& \sqrt{x}\\
y' &=& \frac{d}{dx}x^{\frac{1}{2}} &=& \frac{1}{2} x^{-\frac{3}{2}}&=&\frac{1}{2x\sqrt{x}}\\
y'' &=& \frac{d}{dx}\frac{1}{2}x^{-\frac{3}{2}}&=& -\frac{3}{4}x^{-\frac{5}{2}} &=& -\frac{3}{4x^2\sqrt{x}}\\
y''' &=&-\frac{d}{dx}\frac{3}{4}x^{-\frac{5}{2}}&=&\frac{15}{8} x^{-\frac{7}{2}} &=& \frac{15}{8x^3\sqrt{x}} \\
\end{eqnarray}\]
Example
\[\frac{dy}{dx} = \frac{d}{dx} \left(x^2 + 4x +4\right) =\frac{d}{dx} x^2 + \frac{d}{dx}4x +\frac{d}{dx}4=\] \[=\frac{d}{dx} x^2 + 4\frac{d}{dx}x + 0 = 2x+4\]
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examples
\[y= 4x^2;\quad y' = 4 \frac{d}{dx} x^2= 4(2x)= 8x\] \[y = x^2 x^3 = x^5; \frac{dy}{dx} = 5 x^4= x^2 3x^2 + x^3(2x)= 3x^4 +2x^4 = 5x^4\] \[y = \sqrt{x}; y^\prime = \frac{1}{2x^{1/2}};y'=(-3/2)(1/2)(1/x^{3/2})= -\frac{3x^{\frac{-3}{2}}}{4};y'''=\frac{9x^{\frac{-5}{2}}}{8}\] \[y = (2x +2)^2(x + 1)= 4(x+1)^3= 4x^3 + 12x^2 + 12x + 4\]
Rules about exponents and logs
\[\ln(A B) = \ln(A) + \ln(B);\ \ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B); \ \ln\left(A^b\right) = b\cdot \ln(A);\ A= e^{\ln(A)}\]
\[\frac{d}{dx} \left(e^x\right) = e^x\]
\[\frac{d}{dx} \left(a^x\right) = a^x \ln(a)\] \[\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}\] # e
\[e = \sum_{n=0}^{\infty} \frac{1}{n!}\] \[e =\lim_{n\rightarrow \infty} \left(1 + \frac{1}{n}\right)^n\]
Trigonometry Rules
\[\frac{d}{dx} \left(\sin(x)\right) = cos(x)\]
\[\frac{d}{dx} \left(\cos(x)\right) = -sin(x)\]
Reciprocal rule
\[\frac{d}{dx}\left(\frac{1}{u(x)}\right) = - \frac{\frac{d}{dx}\left(u(x)\right)}{u^2}\]
\[\frac{d}{dx}\left(\frac{1}{x^2}\right)= -\frac{2x}{x^4} = -\frac{2}{x^3}\]
Quotient Rule
\[\frac{d}{dx}\left(\frac{u\left(x\right)}{v\left(x\right)}\right) =\frac{v(x)\cdot\frac{d}{dx}u\left(x\right) -u(x)\cdot\frac{d}{dx}v(x)}{v\left(x\right)^2}\]
\[\begin{eqnarray}\frac{d}{dx} \left(\frac{x+1}{x^2}\right)&=&\frac{x^2\cdot\left(1+0\right) -(x+1)\cdot(2x)}{x^4}\\
&=&\frac{x^2-2x^2-2x}{x^4}\\
&=&-\frac{x^2 +2x}{x^4}=-\frac{x+2}{x^3}\\
\end{eqnarray}\]
Examples
Ex 1: \(\frac{d}{dx} \sin(t^2)\)
Ex 2: \(\frac{d}{dx}\left(\frac{25 x^2}{e^x}\right)\)
Ex 3: \(\frac{d}{dx}\left(\frac{1+e^x}{1+e^{-x}}\right)\)
Ex 4: \(\frac{d}{dx}\left(cos\left(3x^3\right)\right)\)
Ex 5: \(\frac{d}{dx}\left(x^4 cos\left(3x^3\right) sin\left(2x\right)\right)\)
Ex 1
$$\[\begin{eqnarray}
\frac{d}{dx} \sin(t^2)
& = & \cos(u) \frac{du}{dt}\\
& = & \cos\left(t^2\right) 2t\\
&=& 2t \cos\left(t^2\right)
\end{eqnarray}\]$$
Ex 2
\[\begin{eqnarray}
\frac{d}{dx}\left(\frac{25 x^2}{e^x}\right)&=&\frac{e^x\frac{d}{dx}\left(25 x^2\right) -25x^2\cdot e^{x}}{e^{2x}}\\
&=&\frac{e^x\cdot 50x -25x^2\cdot e^{x}}{e^{2x}}\\
&=&\frac{25x\ e^x (2-x)}{e^{2x}} = \frac{25x (2-x)}{e^{x}}\\
\end{eqnarray}\]
Ex 3
$$\[\begin{eqnarray}
\frac{d}{dx}\left(\frac{1+e^x}{1+e^{-x}}\right) & = & \frac{\frac{d}{dx}\left(1+e^{x}\right) \left(1+e^{-x}\right) - \left(1+e^{x}\right)
\frac{d}{dx} \left(1+e^{-x})\right)}{\left(1+e^{-x}\right)^2}\\
& = & \frac{\left(\frac{d}{dx}\left(1\right) +\frac{d}{dx} \left(e^{x}\right)\right) \left(1+e^{-x}\right)}{\left(1+e^{-x}\right)^2} - \\
& & \quad\quad \frac{\left(1+e^{x}\right)
\left(\frac{d}{dx} \left(1\right)+ \frac{d}{dx} \left(e^{-x}\right)\right)}{\left(1+e^{-x}\right)^2}\\
& = & \frac{e^{x} \left(1+e^{-x}\right) - \left(1+e^{x}\right)
\left(e^{-x} \frac{d}{dx} \left(-x\right)\right)}{\left(1+e^{-x}\right)^2}\\
& = & \frac{e^{x} \left(1+e^{-x}\right) + e^{-x}\left(1+e^{x}\right)}{\left(1+e^{-x}\right)^2}\\
\end{eqnarray}\]$$
Ex 4
\[\begin{eqnarray}\frac{d}{dx}\left(cos\left(3x^3\right)\right)& =& -sin(u) \frac{du}{dx}\\
&=& -sin\left(3x^3\right) 9x^2\\
&=& -9x^2 sin\left(3x^3\right)\\
\end{eqnarray}\]
Ex 5
$$\[\begin{eqnarray}
\frac{d}{dx}\left(4x^4\ \cos\left(3x^3\right) \sin\left(2x^2\right)\right) &=& \frac{d}{dx} \left(4x^4\right) \cos\left(3x^3\right) \sin\left(2x^2\right) \\
& &\quad\quad + 4x^4\ \frac{d}{dx}\left(\cos\left(3x^3\right)\right) \sin\left(2x^2\right) \\
& &\quad\quad + 4x^4\ \cos\left(3x^3\right) \frac{d}{dx}\left(\sin\left(2x^2\right)\right) \\
&=& 16x^3 \cos\left(3x^3\right) \sin\left(2x^2\right) \\
& &\quad\quad - 4x^4\ \sin\left(3x^3\right) \frac{d}{dx}\left(3x^3\right) \sin\left(2x^2\right) \\
& &\quad\quad + 4x^4\ \cos\left(3x^3\right) \cos\left(2x^2\right)\frac{d}{dx}\left(2x^2\right) \\
&=& 16x^3 \cos\left(3x^3\right) \sin\left(2x^2\right) \\
& &\quad\quad - 36x^2\ \sin\left(3x^3\right) \sin\left(2x^2\right) \\
& &\quad\quad + 16x^5\ \cos\left(3x^3\right) \cos\left(2x^2\right) \\
\end{eqnarray}\]$$
