Reviewing Calculus: Derivatives - Part 1

Continuing from the review of limits, this post will be the first of a many part review of derivatives. There's so much to cover in derivatives its tough to know what all to include and where to start. I'll try to cover enough that you will be able to take the derivative of just about anything by working/reading through these posts.

I'll begin by stating the definition. A derivative is the slope of the tangent line, of a function; it is the rate of change at a given point in a function. Wikipedia defines it as, a measure of how a function changes as its input changes. The mathematical definition is as follows:$$\lim \limits_{\Delta x \to 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}$$So, let's talk about differentiation.

Given the function $f(x)$, $f'(x)$ is it's derivative. That is to say, $$f'(x) = \lim \limits_{\Delta x \to 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}$$Note: One of the other common derivative notation's - For $y = f(x)$, the derivative can be written as $\frac{dy}{dx}$ or $\frac{d}{dx}(f(x))$

Let's look at a few examples.

Differentiate the functions,  $f(x) = x^2 + 24$ and $g(x) = 4x^3 - 9x + 15$. 

Starting with $f(x)$, we can substitute the function into the limit and evaluate it.$$f'(x) = \lim \limits_{\Delta x \to 0}\frac{((x + \Delta x)^2 + 24) - (x^2 + 24)}{\Delta x}=\lim \limits_{\Delta x \to 0}\frac{x^2 + 2x\Delta x + \Delta x^2 + 24 - x^2 - 24}{\Delta x}$$ $$= \lim \limits_{\Delta x \to 0}\frac{2x\Delta x + \Delta x^2}{\Delta x}= \lim \limits_{\Delta x \to 0}\frac{\Delta x(2x + \Delta x)}{\Delta x}= \lim \limits_{\Delta x \to 0}2x + \Delta x = 2x$$Same plan with $g(x)$. $$g'(x) = \lim \limits_{\Delta x \to 0}\frac{(4(x + \Delta x)^3 - 9(x + \Delta x) + 15) - (4x^3 - 9x + 15)}{\Delta x}$$ $$=\lim \limits_{\Delta x \to 0}\frac{(4(x^3 + 3x^2\Delta x + 3x\Delta x^2 + \Delta x^3) - 9x - 9\Delta x + 15) - 4x^3 + 9x - 15}{\Delta x}$$ $$= \lim \limits_{\Delta x \to 0}\frac{4x^3 + 12x^2\Delta x + 12x\Delta x^2 + 4\Delta x^3 - 9x - 9\Delta x + 15 - 4x^3 + 9x - 15}{\Delta x}$$ $$= \lim \limits_{\Delta x \to 0}\frac{12x^2\Delta x + 12x\Delta x^2 + 4\Delta x^3 - 9\Delta x}{\Delta x}= \lim \limits_{\Delta x \to 0}\frac{\Delta x(12x^2 + 12x\Delta x + 4\Delta x^2 - 9)}{\Delta x}$$ $$= \lim \limits_{\Delta x \to 0}12x^2 + 12x\Delta x + 4\Delta x^2 - 9 = 12x^2 - 9$$We have now successfully differentiated two functions: $f'(x) = 2x$ and $g'(x) = 12x^2 - 9$.

At this point, a pattern may have become apparent. Regardless, shortcuts exist for taking derivatives but, it's important to understand how they work first. Since we've worked a few examples of this style of derivative already, the first one we will be looking at is for functions of the form $f(x) = x^a$. The method for this is as follows, $f'(x) = ax^{a-1}$. This should be clear after a few examples.

Let's differentiate the following: $f(x) = x^7$, $g(x) = 3x^3 + 6$ and $h(x) = -x^2$: $$f'(x) = (7)x^{7-1} = 7x^6$$ $$g'(x) = (3)3x^{3-1} + (0)6^{0-1} = 9x^2$$ $$h'(x) = (2)-x^{2-1} = -2x$$Note: It may be easier to ignore constants in functions when taking derivatives. Since, we are taking the derivative with respect to a variable ($x$ in our examples), parts of the function that don't have that variable can be effectively ignored. For example, when taking the derivative of $f(x) = 3x + 4$, you can ignore the $4$ since it's not affecting the variable.

Hopefully, that was relatively straight-forward. We're going to stop here for now. In the next post, we'll go over adding, subtracting, multiplying and dividing derivatives. Afterward, we'll move on to the trigonometric, logarithmic and exponential derivatives and some application. Eventually, after derivatives have been explained well, we'll move on to some more difficult areas of calculus.

Any questions, comments and/or suggestions are appreciated.