The Chain Rule and Product Rule can be established visually fairly easily 1 and the proofs are fairly straightforward. 2
\[\begin{aligned} \frac{\operatorname{d} }{\operatorname{d} x}\left( u\cdot v \right)&= \frac{\operatorname{d} u}{\operatorname{d} v}\cdot v + u \cdot \frac{\operatorname{d} v}{\operatorname{d} x} \label{rule_11}\\ \ \notag \\ \frac{\operatorname{d} }{\operatorname{d} x}\left( f\left( x \right)\cdot g\left( x \right) \right)&= f'\left( x \right)\cdot g\left( x \right)+ f\left( x \right)\cdot g'\left( x \right) \label{prodruledefleib}\end{aligned}\]
\[\begin{aligned} \frac{\operatorname{d}y }{\operatorname{d} x} &= \frac{\operatorname{d} y}{\operatorname{d} u} \cdot \frac{\operatorname{d} u}{\operatorname{d} x}\\ \ \notag \\ \frac{\operatorname{d} }{\operatorname{d} x}\left[ f\left( g\left( x \right) \right) \right]&= f'\left( g\left( x \right) \right)\cdot g\left( x \right) \\ \ \notag \end{aligned}\]
The chain rule can be used for integration with some clever substitution:
Let:
\[\begin{aligned} \begin{matrix} u &= g(x) & \quad F(x): \enspace F'(x) = f(x) = y \\ \frac{du}{dx} &= g'(x) \end{matrix}\end{aligned}\]
Now by direct substitution into the chain rule:
\[\begin{aligned} \frac{\operatorname{d} }{\operatorname{d} x}\left[ F'\left( u \right) \right]&= F'\left( g\left( x \right) \right)\cdot g'\left( x \right)\notag \\ &= f\left( g\left( x \right) \right)\cdot g'\left( x \right)\notag \\ \implies f\left( g\left( x \right) \right)\cdot g'\left( x \right)&= \frac{\operatorname{d} }{\operatorname{d} x}\left[ F\left( u \right) \right]\notag \\ f\left( g\left( x \right) \right)\cdot g'\left( x \right)&= \frac{\operatorname{d} }{\operatorname{d} x}\left[ F\left( u \right) +C \right]\notag \\ \end{aligned}\]
Now by integrating both sides:
\[\begin{aligned} \int^{}_{} f\left( g\left( x \right) \right)\cdot g'\left( x \right) \operatorname{d}x &= \int^{}_{} \frac{\operatorname{d} }{\operatorname{d} x}\left[ F\left( u \right)+ C \right] \operatorname{d}x \notag \\ &= F\left(u \right) + C\notag \\ &= \int^{}_{} f\left( u \right) \operatorname{d}u\notag \end{aligned}\]
So what we have is integration by substitution:
\[\begin{aligned} \int^{}_{} f\left( g\left( x \right) \right)\cdot g'\left( x \right) \operatorname{d}x &= \int^{}_{} f\left( u \right) \operatorname{d}u \label{ibysub1} \\ \int^{}_{} f\left( u \right)\cdot \frac{\operatorname{d}u }{\operatorname{d} x} \operatorname{d}x&= \int^{}_{} f\left( u \right) \operatorname{d}u \label{ibysubl} \end{aligned}\]
This basically means that if an integral looks like the differentials could cancel out, they do, making the Leibniz notation particularly useful.
The product rule can be used for integration, but it’s only fruitful when:
You can choose some \(u= f\left( x \right)\) that simplifies when differentiated Or atleast stays the same e.g. \(\frac{\operatorname{d} }{\operatorname{d} x}\left[ \sin{\left( x \right)} \right]= \cos{\left( x \right)}\)
\(\operatorname{d} v= g'\left( x \right) \operatorname{d} x\) can be chosen such that the differential can be readily integrated to give \(v\).
Consider the Product Rule \(\left( \ref{prodruledefleib} \right)\):
\[\begin{aligned} \frac{\operatorname{d} }{\operatorname{d} x}\left[ f\left( x \right)\cdot g\left( x \right) \right]&= f'\left( x \right)\cdot g\left( x \right)+ f\left( x \right)\cdot g'\left( x \right)\notag \\ \end{aligned}\]
Let,
\[\begin{aligned} \begin{matrix} &u = f\left( x \right) &&v = g\left( x \right)\\ &\frac{\operatorname{d}u }{\operatorname{d} x}= f'\left( x \right) && \frac{\operatorname{d}v }{\operatorname{d} x} = g'\left( x \right) \end{matrix}\end{aligned}\]
Now we have:
\[\begin{aligned} \int^{}_{} \left( \frac{\operatorname{d}u }{\operatorname{d} x}\cdot v + u\cdot \frac{\operatorname{d}v }{\operatorname{d} x} \right) \operatorname{d}x &= u\cdot v\notag \\ \ \notag \\ \int^{}_{} \left( v\cdot \frac{\operatorname{d}u }{\operatorname{d} x} \right) \operatorname{d}x + \int^{}_{} \left( u\cdot \frac{\operatorname{d}v }{\operatorname{d} x} \right) \operatorname{d}x&= u\cdot v \notag \end{aligned}\]
By Rule \(\left( \ref{ibysubl} \right)\) we have:
\[\begin{aligned} \int^{}_{} v \operatorname{d}u + \int^{}_{} u \operatorname{d}v &= u\cdot v \notag \\ \int^{}_{} u \operatorname{d}v &= u\cdot v - \int^{}_{} v \operatorname{d}u \label{ibypartl} \end{aligned}\]
These are really the only two rules we’ve got (other than manipulation with partial fractions if possible) so the only trick is choosing when to use which one:
Look at the intergrand \(\int^{}_{} \left[\enspace \right] \operatorname{d}x\) :
If it’s of the form\(\left[ f\left( u \right)\cdot \frac{\operatorname{d}u }{\operatorname{d} x}\right] = \left[ f\left( g\left( x \right) \right)\cdot g'\left( x \right) \right]\)
If it’s of the form: \(\left[ f\left( x \right)\cdot \frac{\operatorname{d}u }{\operatorname{d} x} \right]= \left[ f\left( x \right)\cdot g'\left( x \right) \right]\)
Equations involving differentials like \(\operatorname{d} y\) or \(\operatorname{d} x\) or \(\frac{\operatorname{d}y }{\operatorname{d} x}\) are differential equations.
If the derivatives correspond to only a single independent variable and do not involve partials e.g. (\(\frac{\partial u }{\partial x}\)) they are said to be Ordinary Differential Equations (ODE).
The Order of a differential corresponds to the higest derivative taken
The Degree is the highest power of the highest order derivative of the ODE
A Linear ODE is of the form:
\[\begin{aligned} \sum^{n}_{0} \left[ a_0\left( x \right)\cdot \left( \frac{\operatorname{d}^ny }{\operatorname{d} x^n} \right) \right] \label{lindef}\end{aligned}\]
In order of preference, to solve a differential equaiton:
Solve for \(y\) (explicit)
Solve for \(x\) (explicit)
Solve for 0 (implicit)
A differential equation of the form: \[\begin{aligned} g\left( y \right)\cdot \frac{\operatorname{d}y }{\operatorname{d} x} = f\left( x \right) \label{sepdiffform}\end{aligned}\] Is a seperable Ordinary Differential Equation and has a solution:
\[\begin{aligned} \int^{}_{} g\left( y \right) \operatorname{d}y = \int^{}_{} f\left( x \right) \operatorname{d}x \label{sepdiffsol} \end{aligned}\]
\[\begin{aligned} g\left( y \right)\cdot \frac{\operatorname{d}y }{\operatorname{d} x} &= f\left( x \right)\notag \\ \implies \int^{}_{} g\left( y \right)\frac{\operatorname{d}y }{\operatorname{d} x} \operatorname{d}x &= \int^{}_{} f\left( x \right) \operatorname{d}x\notag \\ \end{aligned}\]
By the Substitution Rule at (\(\ref{ibysubl}\)): \[\begin{aligned} \int^{}_{} g\left( y \right) \operatorname{d}y &= \int^{}_{} f\left( x \right) \operatorname{d}x \end{aligned}\]
Some equations can be tricky to deal with, there is a method of \(u\)-Substitution:
Take some equation of the form: \[\frac{\operatorname{d}y }{\operatorname{d} x}= f\left( \frac{x}{y} \right)\] We can perform the \(u\)-substitution:
\[\begin{aligned} u&= \frac{y}{x}\notag \\ \implies y &= u\cdot x\notag \\ \implies \frac{\operatorname{d}y }{\operatorname{d} x}&= \frac{\operatorname{d}u }{\operatorname{d} x}\cdot x+ \left( 1 \right)\cdot u \notag\end{aligned}\]
Substituting in the terms: \[\begin{aligned} \frac{\operatorname{d}y }{\operatorname{d} x}&= f\left( \frac{y}{x} \right)\notag \\ \frac{\operatorname{d}u }{\operatorname{d} x}\cdot x + u&= f\left( u \right)\notag \\ \frac{\operatorname{d}u }{\operatorname{d} x}\cdot x&= f\left( u \right)- u\notag \\ \frac{1}{f\left( u \right)- u}\cdot \frac{\operatorname{d}u }{\operatorname{d} x}\cdot x &= 1\notag \\ \frac{1}{f\left( u \right)- u }\cdot \frac{\operatorname{d}u }{\operatorname{d} x}&= \int^{}_{} \frac{1}{x} \operatorname{d}x \notag \\ \int^{}_{} \frac{1}{f\left( u \right)- u}\cdot \frac{\operatorname{d}u }{\operatorname{d} x} \operatorname{d}x&= \int^{}_{} \frac{1}{x} \operatorname{d}x\notag \\ \int^{}_{} \frac{1}{f\left( u \right)- u} \operatorname{d}u &= \ln{ \left| x \right| }+ c \end{aligned}\]
Now presume \(\exists G\left( u \right): G\left( u \right)= \int^{}_{} \frac{1}{f\left( u \right)- u} \operatorname{d}u\)
\[\begin{aligned} G\left( u \right)&= \ln{ \left| x \right| }+ c\notag \\ G\left( \frac{y}{x} \right)&= \ln{ \left| x \right| }+ c\notag \\ G\left( \frac{y}{x} \right)+ \ln{ \left| x \right| }+ c &= 0 \label{redsepfin}\end{aligned}\]
Hence by by (\[redsepfin\]) there must atleast be an implicit solution to the equation assuming that the integral can be solved.