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# Derivatives (Part III) - Calculus Tutorials

Notation: The derivative $f'(x)$ of a function $f(x)$ is also denoted as

$\frac{df}{dx} (x)$

This notation comes from the fact that when you compute the derivative, you compute

$\frac{df}{dx}(x_0) = \displaystyle\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

The term $f(x)-f(x_0)$ is usually referred as $\Delta f$, and the term $x-x_0$ is referred as $\Delta x$. So, sometimes, in some books (specially Physics books) you're going to find the definition

$\frac{df}{dx}(x_0) = \displaystyle\lim_{\Delta x\to 0} \frac{\Delta f}{\Delta x}$

Theorems to calculate Derivatives

Now it is the time to introduce the heavy artillery. In practice, you won't be computing the limit

$f'(x_0) = \displaystyle\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

very often. It is very important to know how to do it that way, but most of the times it won't be necessary.

Example: Compute the derivative of the function $$f(x)=x^3+x^2$$.

Solution: What do we do here, do we apply the limit to compute the derivative?? Well, your line of reasoning should be the following: The function $f(x) = x^3+x^2$ corresponds to the sum of $x^3$ and $x^2$. The intuition is that if I could compute the derivative of each term \textit{separately}, then I could simplify the calculation.

In other words, if I knew what is the derivative of $x^3$, and if I also knew what is the derivative of $x^2$, then I should know what is the derivative of $x^3+x^2$.....

$\star$ In fact, you do. We have the following theorem:

Theorem: The Derivative of the Sum of Two Functions

Assume that $f(x)$ and $g(x)$ are \textit{differentiable} at $x_0$ (that means that the derivative exists at that point). Then, we have that

$\frac{d}{dx}(f(x)+g(x)) = \frac{df}{dx}(x) + \frac{dg}{dx}(x)$

In other words, the derivative of the sum is the sum of the derivatives (These are not empty words, they really describe the result accurately). This is usually referred as the Linearity Property of the derivative

Now we show a result that will help us to compute a lot of derivatives:

Theorem: The following holds true for all $n\ne 0$:

$\frac{d}{dx} x^n = n x^{n-1}$

Proof: We won't do anything too deep, just to save you from fatal boredom, but let's just do this one to get a feel of it. By definition

$\frac{df}{dx}(x_0) = \displaystyle\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=\displaystyle\lim_{x\to x_0} \frac{x^n- x_0^n}{x-x_0}$ $= \displaystyle\lim_{x\to x_0} \frac{(x- x_0)(x^{n-1}+x^{n-2}x_0 + ...+ xx_0^{n-2} +x_0^{n-1})}{x-x_0} = \displaystyle\lim_{x\to x_0}(x^n+x^{n-1}x_0 + ...+ xx_0^n +x_0^n)$ $= x_0^{n-1}+x_0^{n-2}x_0 + ...+ x_0x_0^{n-2} +x_0^{n-1} = n x_0^{n-1}$

So, let's come back to the problem of finding the derivative of $f(x) = x^3+x^2$. Using the Linearity of derivatives we find that

$\frac{d}{dx}(x^3+x^2) = \frac{d}{dx} x^3 + \frac{d}{dx}x^2 = 3x^2 + 2x$

Let's recall that $\frac{d}{dx}x^n = nx^{n-1}$, so applying that to the case $n=3$ and $n=2$ respectively we get the previous result. The Linearity property can be written in a more general way:

Theorem: Assume that $f(x)$ and $g(x)$ are \textit{differentiable} at $x_0$ and $a$ and $b$ are constants. Then

$\frac{d}{dx}(af(x)+ bg(x)) = a\frac{df}{dx}(x) + b\frac{dg}{dx}(x)$

Below we show an example of how to apply this result:

Example: Compute the derivative of the function $f(x)=3 x^3+2 x^{1/2}$.

Solution: Using Linearity, we get that

$\frac{d}{dx}(3x^3+ 2x^{1/2}) = 3\frac{d}{dx} x^3 + 2 \frac{d}{dx}x^{1/2} = 3(3x^2) + 2\left(\frac{1}{2} x^{-1/2}\right)$ $=9x^2 + x^{-1/2}$

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