Calculation rules for differentials

Differentials and integrals: Differentials

Calculation rules for differentials

The aforementioned relationship between differentials and derivatives makes the following translation of mathematical rules to differentiate rules for calculating differentials possible:

$\begin{aligned} \dd\bigl(c\cdot f(t)\bigr) &= c\cdot \dd\bigl(f(t)\bigr), \text{ for any constant }c\\ \\ \dd\bigl(f(t)+g(t)\bigr) &= \dd\bigl(f(t)\bigr)+\dd\bigl(g(t)\bigr)\;\;\; (\text{sum rule})\\ \\ \dd\bigl(f(t)\cdot g(t)\bigr) &= g(t)\cdot \dd\bigl(f(t)\bigr)+f(t)\cdot \dd\bigl(g(t)\bigr)\;\;\; (\text{product rule})\\ \\ \dd\left(\frac{f(t)}{g(t)}\right) &= \frac{g(t)\cdot \dd\bigl(f(t)\bigr)-f(t)\cdot \dd\bigl(g(t)\bigr)}{\bigl(g(t)\bigr)^2}\;\;\; (\text{quotient rule})\\ \\ \dd\Bigl(f\bigl(g(t)\bigr)\Bigr) &= f'\bigl(g(t)\bigr)\cdot \dd\bigl(g(t)\bigr)= f'\bigl(g(t)\bigr)\cdot g'(t)\,\dd t\;\;\; (\text{chain rule})\\ \end{aligned}$

In a short and more clear notation, the first three calculation rules can be written as:

Shorthand notation of rules for differentials

$\begin{aligned} \dd(c\cdot f) &= c\cdot \dd f, \text{ for any constant }c\\ \\ \dd(f+g) &= \dd f+\dd g\;\;\; (\text{sum rule})\\ \\ \dd(f\cdot g) &= g\cdot \dd f+f\cdot \dd g\;\;\; (\text{product rule})\\ \\ \dd\left(\frac{f}{g}\right) &= \frac{g\cdot \dd f-f\cdot \dd g}{g^2}\;\;\; (\text{quotient rule})\\ \end{aligned}$

We give some examples of working with the calculation rules for differentials.

Write the following differential in the form $f'(y)\,\dd y$ .

$\dd(3y^2-3y+3)=(6y-3)\,\dd y$

Via the constant factor rule and the sum rule for differentials, and via the derivative of a power function, this goes as follows:

$\begin{aligned} \dd(3y^2-3y+3) &= \dd(3y^2)+\dd(-3y)+\dd(3)\\ &\phantom{uvwxyz}\blue{\text{sum rule}}\\ &= 3\,\dd(y^2)-3\,\dd y+\dd(3)\\ &\phantom{uvwxyz}\blue{\text{constant factor rule}}\\ &=3\cdot 2y\,\dd y-3\,\dd y+0\\ &\phantom{uvwxyz}\blue{\text{derivative of a power}}\\ &= 6y\,\dd y-3\,\dd y\\\ &\phantom{uvwxyz}\blue{\text{simplification}}\\ &= (6y-3)\,\dd y\\ &\phantom{uvwxyz}\blue{\text{rewriting}}\end{aligned}$
You can also do the exercise by reading $3y^2-3y+3$ as a function $f(y)$ and calculating the derivative $f'(y)$ . After all, $\dd\bigl(f(y)\bigr) = f'(y)\,\dd y$ .

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