**Step 1:** Simplify and look for algebraic rewrites.

This is still follows are standard derivative process. That means it is the first thing you will *
always
* want to do before you start actually applying a derivative rule. You must ensure that the equation is ready to have a derivative taken.

- Roots or Radicals
- x’s on the bottom of a fraction.
- Trig functions raised to a power.

**Step 2:** Wrap the entire equation in a set of parentheses with a $\frac{d}{d\left(\textcolor[rgb]{}{\mathit{whatever\; your\; respect\; to\; variable}}\right)}$ in the front of the parentheses.

This is mostly just a notational step to let those reviewing (grading) your work that you are applying the implicit differentiation processes. It also helps you to keep track of what “with respect to” process you are applying to the entire equation.

**Step 3:** Look at each term as its own individual math problem, and decide what derivative rule you will need to use on each term.

We break derivative problems into bitesize chunks by the plusses (**
+
**) and minuses (–), and remember that all your derivative rules still apply in these situations. Power rule, product rule, quotient rule, chain rule, and special case rules are still the *
only
*derivative rules you have to work with. If you have an *
x
* and a *
y
* being *
multiplied
* in a single term ($5\textcolor[rgb]{}{x}{\textcolor[rgb]{}{y}}^{\textcolor[rgb]{}{2}}$), then that means you need to apply the product rule to that specific term. Or if you have an *
x
* and a *
y
**
dividing
* each other $\left(\frac{{\textcolor[rgb]{}{x}}^{\textcolor[rgb]{}{3}}}{{\textcolor[rgb]{}{y}}^{\textcolor[rgb]{}{5}}}\right)$, you would need to follow the quotient rule.

**Step 4:** Start performing the derivative rules to each term (bitesize chunk) you identified in *Step 3*.

When doing implicit differentiation, I have a saying I use, “Every time I take a derivative, I get extra stuff.” You will take the derivative of each bitesize chunk like you normally would, but each time you take a derivative you will get an extra piece.

What that extra piece will look like is $\frac{d\overline{)\textcolor[rgb]{}{\mathit{the}}\mathrm{}\textcolor[rgb]{}{\mathit{letter}}\mathrm{}\textcolor[rgb]{}{\mathit{of}}\mathrm{}\textcolor[rgb]{}{\mathit{what}}\mathrm{}\textcolor[rgb]{}{\mathit{you}}\mathrm{}\textcolor[rgb]{}{\mathit{took}}\mathrm{}\textcolor[rgb]{}{\mathit{the}}\mathrm{}\textcolor[rgb]{}{\mathit{derivative}}\mathrm{}\textcolor[rgb]{}{\mathit{of}}}}{\textcolor[rgb]{}{d}\left(\textcolor[rgb]{}{\mathit{whatever}}\textcolor[rgb]{}{\mathrm{}}\textcolor[rgb]{}{\mathit{your}}\textcolor[rgb]{}{\mathrm{}}\textcolor[rgb]{}{\mathit{respect}}\textcolor[rgb]{}{\mathrm{}}\textcolor[rgb]{}{\mathit{to}}\textcolor[rgb]{}{\mathrm{}}\textcolor[rgb]{}{\mathit{variable}}\right)}$.

If you were taking the derivative of a *
y
* “with respect to” *
x
*, you would get $\frac{d\overline{)\textcolor[rgb]{}{y}}}{d\overline{)\textcolor[rgb]{}{x}}}$ as your extra stuff. If you were taking the derivative of an *
x
* “with respect to” *
x
*, you would get $\frac{d\overline{)\textcolor[rgb]{}{x}}}{d\overline{)\textcolor[rgb]{}{x}}}$ as your extra stuff.

**Step 5:** Clean up the equation.

After you have applied all of your derivative rules in *Step 4*, you will want to look to clean up the equation you are now working with. You are about to transition from doing *calculus* into doing some *algebra* to complete this process. That makes this is a great point in the process to make sure all the terms make sense, and are ready for *algebra*. You work your way left to right through the derivative equation from *Step 4*, making sure that each term is ready for some *algebra*.

There are three main clean up moves I look to do:

1) Cancel and combine

- Combine like terms through addition and subtraction.
- Cancel out pieces that are being divided by themselves.

The most common cancelation in implicit differentiation is when you get $\frac{\mathit{dx}}{\mathit{dx}}$. It is just like having $\frac{5}{5}$, the 5’s cancel each other out, and you are left with 1. The same is true with $\frac{\mathit{dx}}{\mathit{dx}}$ the $\mathit{dx}$’s would cancel each other out and you would be left with 1.

2) Rewrites

The most common rewrite is *negative exponents*. I always tell people *negative exponents* are great for *calculus*, but they suck for *algebra*. Make sure you do the rewrites necessary to make all your *exponents positive*.

3) Proper order

You just want to take the time to make sure each term is written in the “proper form”. Generally, this is numbers then alphabetical order of your variables with your derivative pieces ($\frac{\mathit{dy}}{\mathit{dx}}\mathit{or}\frac{\mathit{dx}}{\mathit{dx}}$) at the end of the term. It is a little move, but one that will really help your brain see all the terms better.

**Step 6:** Solve for your derivative piece. (i.e., your $\frac{\mathit{dy}}{\mathit{dx}},\frac{\mathit{dy}}{\mathit{dt}}$)

You will need to do the *algebra* necessary to get your derivative piece (i.e., your $\frac{\mathit{dy}}{\mathit{dx}},\frac{\mathit{dy}}{\mathit{dt}}$) alone on one side of your equal sign. One of the most common situations that you will find yourself in with these problems, is that you have multiples of what you are trying to solve for in the equation. You might have $\frac{\mathit{dy}}{\mathit{dx}}$’s in three or four terms of your equation. In order to solve for $\frac{\mathit{dy}}{\mathit{dx}}$ you need to get it down to just one, and it needs to be alone on one side of the equals. The method you will want to use will be to get every term that has what you are looking for on one side of the equals, and every term that doesn’t onto the other side of the equals. This will then allow you to factor out the $\frac{\mathit{dy}}{\mathit{dx}},\frac{\mathit{dy}}{\mathit{dt}}$, or whatever you are looking for, and get it down to just one instances in the equation.

**Possible Step 7 (Find the ****
slope
****of the ****
tangent
****line):** Plug the variables into the derivative equation from *Step 6*.

You may be asked to find the derivative at a specific point, $\left(\textcolor[rgb]{}{x}\textcolor[rgb]{}{,}\textcolor[rgb]{}{y}\right)$, the slope of the tangent line at a specific point $\left(\textcolor[rgb]{}{x}\textcolor[rgb]{}{,}\textcolor[rgb]{}{y}\right)$, or most related rates word problems want some type of final value for the answer. You will need to plug your values into the derivative to find your final answer. Sometimes you will only be given one of the values from the point, $\left(\textcolor[rgb]{}{x}\textcolor[rgb]{}{,}\textcolor[rgb]{}{y}\right)$, but not the other. You are given only the *
x-value
*or the *
y-value
*, but you need both of those values to find your final answer. Keep in mind that the original equation you are given, no matter how messy looking, is still the equation that all your points, $\left(\textcolor[rgb]{}{x}\textcolor[rgb]{}{,}\textcolor[rgb]{}{y}\right)$, come from. Which means you can always plug in the *
x-value
*to get the *
y-value
*or vice versa.

**Possible Step 7 (Find the second derivative, **$\frac{{\mathit{d}}^{\mathit{2}}\mathit{y}}{\mathit{d}{\mathit{x}}^{\mathit{2}}}$**): **Take the derivative, *again,*of your derivative from *Step 6* following the same steps.

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