differentiation rules

Let \[f,g\] be functions. If \[f(x)=g(x)\] for every \[x\in\mathbb{R}\], only then \[f^{\prime}(x)=g^{\prime}(x)\] will hold.

• The derivative of \[h(x)=af(x)+bg(x)\] is \[h^{\prime}(x)=af^{\prime}(x)+bg^{\prime}(x)\]
  
• \[(af)^{\prime}=af^{\prime}\]
  
• \[(f+g)^{\prime}=f^{\prime}+g^{\prime}\]
  
• \[(f-g)^{\prime}=f^{\prime}-g^{\prime}\]
  

\[(u\cdot v)^{\prime}=u^{\prime}\cdot v+u\cdot v^{\prime}\]

Let \[h(x)=f(g(x))\], then \[h^{\prime}=f^{\prime}(g(x))\cdot g^{\prime}(x)\], or in Leibniz notation, \[\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}\].

Let \[f\] be a function, if \[f\] has an inverse function \[f^{-1}(x)\], meaning that \[f^{-1}(f(x))=x\] and \[f(f^{-1}(x))=x\], \[(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}\].

Let \[f(x)=x^{r}\], thus \[f^{\prime}(x)=rx^{r-1}\].

\[\left( \frac{1}{f(x)} \right)^{\prime}=-\frac{f^{\prime}(x)}{(f(x))^{2}}\], or written as \[\frac{d\displaystyle\left(\frac{1}{f}\right)}{dx}=-\frac{1}{f^{2}}\frac{df}{dx}\].

If \[f\] and \[g\] are functions, then \[\left( \frac{f}{g} \right)^{\prime}=\frac{f^{\prime}g-g^{\prime}f}{g^{2}}\].

The chain rule states that \[(f(g(x)))^{\prime}=f^{\prime}(g(x))\cdot g^{\prime}(x)\], suppose now \[y\] is related to \[x\] by \[y=g(x)\]. We can rewrite this as \[(f(y))^{\prime}=f^{\prime}(y)\cdot y^{\prime}\], or the more similar form \[\frac{d}{dx}f(y)=f^{\prime}(y)\cdot \frac{dy}{dx}\].

Let \[x(t)\] and \[y(t)\] be the coordinates of the points of the curve expressed as functions of a variable \[t\]: \[y=y(t)\] and \[x=x(t)\]. Let \[y=f(x)\], \[y(t)=f(x(t))\], using chain rule, \[y^{\prime}(t)=f^{\prime}(x(t))\cdot x^{\prime}(t)\], thus \[f^{\prime}(x(t))=\frac{dy}{dx}=\frac{y^{\prime}(t)}{x^{\prime}(t)}\].