Derivative formulas

• [math]\displaystyle{ f(x) = c }[/math]. This is the most trivial derivative: [math]\displaystyle{ \lim_{x \ \to \ h} \frac{f(x + h) - f(x)}{h} = \frac{c - c}{h} = 0 }[/math]. A constant function never changes its value. Therefore its rate of change is zero everywhere.

• [math]\displaystyle{ f(x) = x^n \implies f'(x) = nx^{n \ - \ 1} }[/math]. One confusion that happens here is caused by the tangent line problem. When we differentiate a second degree polynomial, the resulting function is a first degree polynomial and a straight line. However, when the degree is 3 or higher, we still have the tangent line problem, but the derivative won't be a polynomial of degree equal to one!
  
  [math]\displaystyle{ f(x) = x^{-n} \implies f'(x) = -nx^{-n \ - \ 1} }[/math].
  
  [math]\displaystyle{ f(x) = x^{\frac{1}{n}} \implies f'(x) = \frac{1}{n}x^{n \ - \ \frac{1}{n}} }[/math].

• [math]\displaystyle{ (f \pm g)'(x) = f'(x) \pm g'(x) }[/math]. The derivative of the sum is the sum of the derivatives. The proof is identical to the sum of limits, because a derivative is a limit.

• [math]\displaystyle{ (c \cdot f)'(x) = c \cdot f'(x) }[/math]. The same rule for limits.

• [math]\displaystyle{ (f \cdot g)'(x) = f'(x)g(x) + f(x)g'(x) }[/math]. The derivative of the product is not the product of the derivatives! This is contrary to the previous two. The reason for this is because when we calculate limits we can do the product of the limits because the limit is a fixed point. With derivatives, however, we have rates of change, not a fixed point. The geometrical interpretation is similar to that of [math]\displaystyle{ (a + b)^2 }[/math]:
  
  
  
  Let's call [math]\displaystyle{ f(x) = u }[/math] and [math]\displaystyle{ g(x) = v }[/math]. Let's assume that their respective values are positive and interpret the product as an area of a rectangle.
  
  Note: Can we do [math]\displaystyle{ x^5 = x^3 x^2 }[/math]? Yes, we can use the properties of powers to see one function as a product of two functions. But that won't help in solving derivatives and will only make calculations take unnecessarily longer to perform.