Removing multiplicative inverses from numerators

Question

Recently I've been working on learning Calculus I using the Great Courses Wondrium course "Understanding Calculus: Problems, Solutions, and Tips."

While I'm having no problems understanding the concepts being taught, there's one thing that's boggling me.

You see, the teacher likes to use functions of the form $\frac{1}{x^n}$, where $n$ is a positive integer, for practice problems. When these are used in derivatives they result in something like this:

$$ \frac{d}{dx}[f(x)=\frac{1}{x^2}] = \lim_{\Delta x\to0} \frac{\frac{1}{(x+\Delta x)^2}-\frac{1}{x^2}}{\Delta x} $$

The complex fractions are really boggling me and everybody I have asked for help. According to the answers in the back of the book, the next step in simplification is

$$ \lim_{\Delta x\to0} \frac{x^2-(x+\Delta x)^2}{\Delta x(x+\Delta x)^2x^2} $$

However, it doesn't show how this simplification is actually done.

How do I get from point A to point B?

Answer 1

Your expression (inside the limit) has the form $\frac{\frac1a-\frac1b}{c}$. One way to rewrite this is to multiply top and bottom of the main fraction by something that will clear all the inner denominators--in this case $ab$.

Thus: $$ \frac{\frac1a-\frac1b}{c} = \frac{(\frac1a-\frac1b)\cdot ab}{c \cdot ab}=\frac{b-a}{abc} $$

In your specific situation, $a$ is playing the role of $(x+\Delta x)^2$; $b$ is $x^2$; and $c$ is $\Delta x$.

Answer 2

$$\frac{x^2-(x+\Delta x)^2}{\Delta x(x+\Delta x)^2x^2}=\frac{x^2-x^2-2x\Delta x-(\Delta x)^2}{\Delta x(x+\Delta x)^2x^2}=-\frac{2x\Delta x+(\Delta x)^2}{\Delta x(x+\Delta x)^2x^2}$$ $$=-\frac{\Delta x}{\Delta x}\frac{2x+\Delta x}{x^2(x+\Delta x)^2}=-\frac{2x+\Delta x}{x^2(x+\Delta x)^2}$$ You can the first simplify whith $\Delta x$ because $\Delta x\neq0$ and then get $\Delta x=0$ at limit. $$\lim_{\Delta x\rightarrow0}-\frac{2x+\Delta x}{x^2(x+\Delta x)^2}=-2\frac{x}{x^4}=-2\frac{1}{x^3}$$