Properties of Derivatives

by Justin Skycak (x.com/justinskycak) on

Given a sum, we can differentiate each term individually. But why are we able to do this? Does multiplication work the same way? What about division?

This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Properties of Derivatives. In Justin Math: Calculus. https://justinmath.com/properties-of-derivatives/

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We know that when differentiating polynomials, we can differentiate each term individually. But why are we able to do this? Does multiplication work the same way? What about division? We answer these questions in this chapter.

Sum Rule

First of all, we are able to differentiate each term in a polynomial individually, because in general, derivatives can be separated over addition. The derivative of a sum, is the sum of derivatives of individual terms.

To see why this is true, we can look at what happens in the difference quotient when we take the derivative of the sum of two functions. We are able to rearrange the difference quotient into the sum of difference quotients of the two functions, which shows that the derivative of the sum is just the sum of the derivatives.

$\begin{align*} (f(x)+g(x))' &= \lim\limits_{\Delta x \to 0} \frac{f(x+\Delta x)+g(x+\Delta)-( f(x)+g(x) ) }{\Delta x} \\ &= \lim\limits_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x) +g(x+\Delta)-g(x) }{\Delta x} \\ &= \lim\limits_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x) }{\Delta x} + \lim\limits_{\Delta x \to 0} \frac{g(x+\Delta)-g(x) }{\Delta x} \\ &= f'(x) + g'(x) \end{align*}$


Constant Multiple Rule

Another useful property of derivatives is that constants can be moved outside the derivative.

$\begin{align*} (2x^3)' = 2(x^3)' = 2(3x^2) = 6x^2 \end{align*}$


Combining this with the power rule, we can differentiate entire polynomial expressions.

$\begin{align*} (3x^4+x^2-2x+1)' &= (3x^4)' + (x^2)' + (-2x)' + (1)' \\ &= 3(x^4)' + (x^2)' -2(x)' + (1)' \\ &= 3(4x^3) + (2x) - 2(1) + 0 \\ &= 12x^3 + 2x - 2 \end{align*}$


To see why we can move constants outside the derivative, we can inspect what happens in the difference quotient when we take the derivative of a function multiplied by a constant. The constant factors out, and we can write the result as the product of the constant and the derivative.

$\begin{align*} (cf(x))' &= \lim\limits_{\Delta x \to 0} \frac{cf(x+\Delta x) - cf(x)}{\Delta x} \\ &= \lim\limits_{\Delta x \to 0} c\frac{f(x+\Delta x) - f(x)}{\Delta x} \\ &= c \lim\limits_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} \\ &= cf'(x) \end{align*}$


Product Rule

Taking the derivative of a product, perhaps surprisingly, results in a sum. For each term that is multiplied in the product, a copy of the product is added in the sum, with the particular term replaced by its derivative.

$\begin{align*} \left[ (3x^2+2)(x+3)^2(2x+1)^5 \right]' &= (3x^2+2)'(x+3)^2(2x+1)^5 \\& \hspace{0.5cm} + (3x^2+2)\left[ (x+3)^2 \right]' (2x+1)^5 \\& \hspace{0.5cm} + (3x^2+2)(x+3)^2 \left[ (2x+1)^5 \right]' \\ &= (6x)(x+3)^2(2x+1)^5 \\ &\hspace{.5cm} + (3x^2+2)\left[ 2(x+3) \right] (2x+1)^5 \\ &\hspace{.5cm} + (3x^2+2)(x+3)^2 \left[ 10(2x+1)^4 \right] \\ &= (x+3)(2x+1)^4 \left[ \begin{matrix} 6x(x+3)(2x+1) \\ + (3x^2+2)(2)(2x+1) \\ + (3x^2+2)(x+3)(10) \end{matrix} \right] \\ &= (x+3)(2x+1)^4(54x^3+138x^2+46x+64) \end{align*}$


To see why this works, we can look at what happens in the difference quotient when we take the derivative of the product of two functions. We are able to rearrange the difference quotient into the sum of the difference quotients of the two functions, with each difference quotient multiplied by the other function. This shows that the derivative of the product is a sum of copies of the product, each with one particular term replaced by its derivative.

$\begin{align*} (f(x)g(x))' &= \lim\limits_{\Delta x \to 0} \frac{f(x+\Delta x)g(x+\Delta x) - f(x)g(x)}{\Delta x} \\ &= \lim\limits_{\Delta x \to 0} \frac{ f(x+\Delta x)\left[ g(x+\Delta x) - g(x) \right] + g(x) \left[ f(x+\Delta x)-f(x) \right] }{\Delta x} \\ &= \lim\limits_{\Delta x \to 0} f(x+\Delta x) \frac{g(x+\Delta x) - g(x)}{\Delta x} + \lim\limits_{\Delta x \to 0} g(x) \frac{f(x+\Delta x) - f(x)}{\Delta x} \\ &= f(x+0) \lim\limits_{\Delta x \to 0} \frac{g(x+\Delta x) - g(x)}{\Delta x} + g(x) \lim\limits_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} \\ &= f(x)g'(x) + g(x)f'(x) \\ &= f'(x)g(x) + f(x)g'(x) \end{align*}$


Quotient Rule

To take the derivative of a quotient, we can use the product rule in conjunction with the power rule and chain rule.

$\begin{align*} \left( \frac{f(x)}{g(x)} \right)' &= \left( f(x)g(x)^{-1} \right)' \\ &= f'(x)g(x)^{-1} + f(x)\left( g(x)^{-1} \right)' \\ &= f'(x)g(x)^{-1} + f(x)\left( -g(x)^{-2}g'(x) \right) \\ &= \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{g(x)^2} \\ &= \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} \end{align*}$


Applying this formula can save us the work of combining fractions after differentiating.

$\begin{align*} \left( \frac{x^2}{2x+1} \right)' &= \frac{(x^2)'(2x+1)-(x^2)(2x+1)'}{(2x+1)^2} \\ &= \frac{(2x)(2x+1)-(x^2)(2)}{(2x+1)^2} \\ &= \frac{2x^2+2x}{(2x+1)^2} \end{align*}$


Exercises

Use the properties of derivatives to differentiate the following functions. (You can view the solution by clicking on the problem.)

$\begin{align*}1) \hspace{.5cm} f(x)= \sqrt{x} (x+2)^2 \end{align*}$
Solution:
$\begin{align*} f'(x)= \frac{5x^2+12x+4}{2 \sqrt{x} } \end{align*}$

$\begin{align*}2) \hspace{.5cm} f(x)= (x-3)^2(x+1)^3 \end{align*}$
Solution:
$\begin{align*} f'(x)= (x-3)(x+1)^2(5x-7) \end{align*}$

$\begin{align*}3) \hspace{.5cm} f(x)= x^2(x+1)^3 \end{align*}$
Solution:
$\begin{align*} f'(x)= x(x+1)^2(5x+2) \end{align*}$

$\begin{align*}4) \hspace{.5cm} f(x)= (2x+3)^4(x^2-5)^3 \sqrt{x} \end{align*}$
Solution:
$\begin{align*} f'(x)= \frac{3}{2\sqrt{x} } (2x+3)^3(x-5)^2(14x^3+13x^2-30x-5) \end{align*}$

$\begin{align*}5) \hspace{.5cm} f(x)= \frac{x+1}{x+2} \end{align*}$
Solution:
$\begin{align*} f'(x)= \frac{1}{(x+2)^2} \end{align*}$

$\begin{align*}6) \hspace{.5cm} f(x)= \frac{x}{x^2+1} \end{align*}$
Solution:
$\begin{align*} f'(x)= \frac{1-x^2}{(x^2+1)^2} \end{align*}$

$\begin{align*}7) \hspace{.5cm} f(x)= \frac{x^2+2x+3}{2x^2-5} \end{align*}$
Solution:
$\begin{align*} f'(x)= - \frac{6x^2+22x+10}{(2x^2-5)^2} \end{align*}$

$\begin{align*}8)\hspace{.5cm} f(x) = \frac{x^3-2x}{x^4-1} \end{align*}$
Solution:
$\begin{align*} f'(x)= \frac{-x^6+6x^4-3x^2+2}{(x^4-1)^2} \end{align*}$

$\begin{align*}9) \hspace{.5cm} f(x) = \frac{x^3(x+1)^4}{(x-1)^3} \end{align*}$
Solution:
$\begin{align*} f'(x)= \frac{x^2(x+1)^3 (4x^2-7x-3)}{(x-1)^4} \end{align*}$

$\begin{align*}10) f(x) = \frac{8x^2-\sqrt{x} }{ \sqrt{x} + 2 } \end{align*}$
Solution:
$\begin{align*} f'(x) = \frac{ 32x^\frac{3}{2} + 12x^2 - 1 }{ \left( \sqrt{x}+2 \right)^2 \sqrt{x} } \end{align*}$


This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Properties of Derivatives. In Justin Math: Calculus. https://justinmath.com/properties-of-derivatives/


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