Multiplication by Constant
A constant factor remains unchanged when differentiating.
$$ \begin{aligned}
y &= C \cdot u(t) \\[10pt]
\Rightarrow \qquad \dot{y} &= C \cdot \dot{u}(t) = C \cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u(t)\Bigr]
\end{aligned} $$
Examples:
$$ \begin{alignat}{7}
f(x) &= y = 6x^2 \qquad && \Rightarrow \qquad y^{\prime} &&= 6 \cdot \bigl(x^2\bigr)^{\prime} &&= 6 \cdot 2x = 12x\\[10pt]
f(t) &= y = 3\sin(t) \qquad && \Rightarrow \qquad \dot{y} &&= 3 \cdot \dot{\bigl(\sin(t)\bigr)} &&= 3 \cos(t)
\end{alignat} $$
Sum Rule
A finite sum of functions can be differentiated term by term.
$$ \begin{aligned}
y &= u_1(t) + u_2(t) + \cdots + u_n(t) \\[10pt]
\Rightarrow \qquad \dot{y} &= \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u_1(t)\Bigr] + \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u_2(t)\Bigr] + \cdots + \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u_n(t)\Bigr]
\end{aligned} $$
Examples:
$$
\definecolor{lsgreen}{RGB}{79,175,152}
\definecolor{lsblue}{RGB}{16,160,205}
\definecolor{lsyellow}{RGB}{255,182,0}
\begin{alignat}{7}
f(x) &= y = {\color{red}x^4} + {\color{lsblue}x^3} - {\color{lsyellow}x^2} \qquad && \Rightarrow \qquad y^{\prime} &&= {\color{red}\dfrac{\mathrm{d}}{\mathrm{d}x}\Bigl[x^4\Bigr]} + {\color{lsblue}\dfrac{\mathrm{d}}{\mathrm{d}x}\Bigl[x^3\Bigr]} - {\color{lsyellow}\dfrac{\mathrm{d}}{\mathrm{d}x}\Bigl[x^2\Bigr]} &&= {\color{red}4x^3}+{\color{lsblue}3x^2}-{\color{lsyellow}2x}\\[10pt]
f(t) &= y = {\color{red}6t^2} + {\color{lsblue}3\sin(t)} \qquad && \Rightarrow \qquad \dot{y} &&= {\color{red}\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[6t^2\Bigr]} + {\color{lsblue}\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[3\sin(t)\Bigr]} &&= {\color{red}12t}+{\color{lsblue}3 \cos(t)}
\end{alignat} $$
Product Rule
The product rule can be used to calculate derivatives of functions that are expressed as a product of other functions.
For two factor functions:
$$ \begin{aligned}
y &= u(t) \cdot v(t) \\[10pt]
\Rightarrow \qquad \dot{y} &= \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u(t)\Bigr] \cdot v(t) + u(t) \cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[v(t)\Bigr]
\end{aligned} $$
For three factor functions:
$$ \begin{aligned}
y &= u(t) \cdot v(t) \cdot w(t)\\[10pt]
\Rightarrow \qquad \dot{y} &= \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u(t)\Bigr] \cdot v(t) \cdot w(t) + u(t) \cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[v(t)\Bigr] \cdot w(t) + u(t) \cdot v(t) \cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[w(t)\Bigr]
\end{aligned} $$
Examples:
$$
\definecolor{lsgreen}{RGB}{79,175,152}
\definecolor{lsblue}{RGB}{16,160,205}
\definecolor{lsyellow}{RGB}{255,182,0}
\begin{alignat}{7}
f(t) = y &= {\color{red}t^2} \cdot {\color{lsblue}\sin(t)} \\[7pt]
\Rightarrow \qquad \dot{y} &= {\color{red}\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[t^2\Bigr]} \cdot {\color{lsblue}\sin(t)} + {\color{red}t^2} \cdot {\color{lsblue}\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[\sin(t)\Bigr]} = {\color{red}2t} \cdot {\color{lsblue}\sin(t)} + {\color{red}t^2} \cdot {\color{lsblue}\cos(t)}\\[12pt]
f(x) = y &= {\color{red}x^2} \cdot {\color{lsblue}e^x} \cdot {\color{lsyellow}\sin(x)}\\[7pt]
\Rightarrow \qquad y^{\prime} &= {\color{red}\dfrac{\mathrm{d}}{\mathrm{d}x}\Bigl[x^2\Bigr]} \cdot {\color{lsblue}e^x} \cdot {\color{lsyellow}\sin(x)} + {\color{red}x^2} \cdot {\color{lsblue}\dfrac{\mathrm{d}}{\mathrm{d}x}\Bigl[e^x\Bigr]} \cdot {\color{lsyellow}\sin(x)} + {\color{red}x^2} \cdot {\color{lsblue}e^x} \cdot {\color{lsyellow}\dfrac{\mathrm{d}}{\mathrm{d}x}\Bigl[\sin(x)\Bigr]}\\[7pt]
&= {\color{red}2x} \cdot {\color{lsblue}e^x} \cdot {\color{lsyellow}\sin(x)} + {\color{red}x^2} \cdot {\color{lsblue}e^x} \cdot {\color{lsyellow}\sin(x)} + {\color{red}x^2} \cdot {\color{lsblue}e^x} \cdot {\color{lsyellow}\cos(x)} = {\color{lsblue}e^x} \cdot {\color{red}x} \cdot \bigl({\color{red}2} \cdot {\color{lsyellow}\sin(x)} + {\color{red}x} \cdot {\color{lsyellow}\sin(x)} + {\color{red}x} \cdot {\color{lsyellow}\cos(x)}\bigr)
\end{alignat} $$
Quotient Rule
The quotient rule can be used to calculate derivatives of functions that are represented as quotients of other functions.
$$ \begin{aligned}
y &= \dfrac{u(t)}{v(t)} \\[10pt]
\Rightarrow \qquad \dot{y} &= \dfrac{\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u(t)\Bigr] \cdot v(t) - u(t) \cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[v(t)\Bigr]}{\bigl[v(t)\bigr]^2}
\end{aligned} $$
Examples:
$$
\definecolor{lsgreen}{RGB}{79,175,152}
\definecolor{lsblue}{RGB}{16,160,205}
\definecolor{lsyellow}{RGB}{255,182,0}
\begin{aligned}
f(t) = y &= \dfrac{{\color{red}e^t}}{{\color{lsblue}t^2}} \\[7pt]
\Rightarrow \qquad \dot{y} &= \dfrac{{\color{red}\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[e^t\Bigr]} \cdot {\color{lsblue}t^2} - {\color{red}e^t} \cdot {\color{lsblue}\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[t^2\Bigr]}}{\bigl[{\color{lsblue}t^2}\bigr]^2} = \dfrac{ {\color{red}e^t} \cdot {\color{lsblue}t^2} - {\color{red}e^t} \cdot {\color{lsblue}2t}} {{\color{lsblue}t^4}}= \dfrac{ ({\color{lsblue}t} - {\color{lsblue}2}) \cdot {\color{red}e^t} } {{\color{lsblue}t^3}}\\[12pt]
f(t) = y &= {\color{lsblue}e^{-t}}\cdot {\color{red}\sin(t)} \\[7pt]
\Rightarrow \qquad \dot{y} &= \dfrac{\mathrm{d}}{\mathrm{d}t}\biggl[ \dfrac{{\color{red}\sin(t)}} {{\color{lsblue}e^t}}\biggr] =
\dfrac{{\color{red}\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[\sin(t)\Bigr]} \cdot {\color{lsblue}e^t} - {\color{red}\sin(t)} \cdot {\color{lsblue}\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[e^t\Bigr]}}{\bigl[{\color{lsblue}e^t}\bigr]^2} =
\dfrac{ {\color{red}\cos(t)} \cdot {\color{lsblue}e^t} - {\color{red}\sin(t)} \cdot {\color{lsblue}e^t}} {{\color{lsblue}e^{2t}}}=
\dfrac{ ({\color{red}\cos(t)} - {\color{red}\sin(t)}) \cdot {\color{lsblue}e^t} } {{\color{lsblue}e^{2t}}}\\[7pt]
&= \dfrac{ {\color{red}\cos(t)} - {\color{red}\sin(t)} } {{\color{lsblue}e^t}} = ( {\color{red}\cos(t)} - {\color{red}\sin(t)}) \cdot {\color{lsblue}e^{-t}}
\end{aligned} $$
Chain Rule
The chain rule can be used to calculate the derivative of a function that is represented as a combination of several functions.
$$ \begin{aligned}
y &= f\bigl(u(t)\bigr) \\[10pt]
\Rightarrow \qquad \dot{y} &= \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[f(u)\Bigr] \cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u(t)\Bigr]
\end{aligned} $$
The derivative of a composite (chained) function \(y=f\bigl(u(t)\bigr)\), where \(f(u)\) and \(u(t)\) are the component functions, is the product of the outer and the inner derivative.
Procedure:
- Substitute \(u=u(t)\)
- Take the derivative \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[f(u)\bigr]\) (outer derivative)
- Eliminate the auxiliary variable \(u\) by substituting back
- Take the derivative \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[u(t)\bigr]\) (inner derivative)
- Multiply both derivatives together
Example:
\(f(t) = y = \ln(1+t^4) \)
- \(u = 1+t^4\)
- \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[\ln(u)\bigr] = \frac{1}{u}\)
- \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[\ln(u)\bigr] = \frac{1}{1+t^4}\)
- \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[1+t^4\bigr] = 4t^3\)
- \(\dot{y} = \frac{1}{1+t^4} \cdot 4t^3 = \frac{4t^3}{1+t^4}\)
Chain Rule for Doubly Nested Functions
$$ \begin{aligned}
y &= f\Bigl(v\bigl(u(t)\bigr)\Bigr) \\[10pt]
\Rightarrow \qquad \dot{y} &= \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[f(v)\Bigr] \cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[v(u)\Bigr] \cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[u(t)\Bigr]
\end{aligned} $$
Procedure:
- Substitute \(u=u(t)\)
- Substitute \(v=v(u)\)
- Compute the derivative \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[f(v)\bigr]\) (outer derivative)
- Compute the derivative \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[v(u)\bigr]\) (1st inner derivative)
- Compute the derivative \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[u(t)\bigr]\) (2nd inner derivative)
- Multiply the three derivatives together
- Perform the reverse substitution in the order \(v \rightarrow u \rightarrow t\)
Example:
\(f(t) = y = \sin^3(t^2+t) \)
- \(u = t^2+t \quad \Rightarrow \quad y = \sin^3(u) = (\sin(u))^3\)
- \(v = \sin(u) \quad \Rightarrow \quad y = v^3\)
- \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[v^3\bigr] = 3v^2\)
- \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[\sin(u)\bigr] = \cos(u)\)
- \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[t^2+t\bigr] = 2t+1\)
- \(\dot{y} = 3v^2 \cdot \cos(u) \cdot (2t+1)\)
- \(\dot{y} = 3\sin^2(u) \cdot \cos(u) \cdot (2t+1) = 3(2t+1) \cdot \sin^2(t^2+t) \cdot \cos(t^2+t)\)
Logarithmic Differentiation
Logarithmic differentiation is a technique in differential calculus where functions are transformed into logarithmic form to calculate their derivatives. This can be useful when a function is expressed in exponential form and difficult to manipulate.
$$ \begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}t}\Bigl[\ln(y)\Bigr] &= \dfrac{1}{y}\cdot \dot{y}~
\end{aligned} $$
In logarithmic differentiation, the function \(y=f(t)\) is first logarithmized on both sides and then differentiated using the chain rule.
Logarithmic differentiation is applied to functions of the form \(y= \bigl[u(t)\bigr]^{v(t)}\) with \(u(t)>0\).
Example:
\(f(t) = y = t^{\cos(t)}\quad,t>0 \)
- Logarithmizing: \(\ln(y) = \ln\bigl(t^{\cos(t)}\bigr) = \cos(t) \cdot \ln(t)\)
- Differentiating: \(\frac{\mathrm{d}}{\mathrm{d}t}\bigl[\ln(y)\bigr] = \frac{\mathrm{d}}{\mathrm{d}t}\bigl[\cos(t) \cdot \ln(t)\bigr]\)
- \(\frac{1}{y} \cdot \dot{y} = \frac{\mathrm{d}}{\mathrm{d}t}\bigl[\cos(t)\bigr] \cdot \ln(t) + \cos(t) \cdot \frac{\mathrm{d}}{\mathrm{d}t}\bigl[\ln(t)\bigr]\)
- \(\frac{1}{y} \cdot \dot{y} = - \sin(t) \cdot \ln(t) + \cos(t) \cdot \frac{1}{t}\)
- \(\dot{y} = y \cdot \Bigl(- \sin(t) \cdot \ln(t) + \frac{\cos(t)}{t}\Bigr)\)
- \(\dot{y} = t^{\cos(t)} \cdot \Bigl(- \sin(t) \cdot \ln(t) + \frac{\cos(t)}{t}\Bigr)\)
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