Example 1: Finding the 1st Derivative

Find $\frac{\mathit{dy}}{\mathit{dx}}$ given: $x^3+y^2=x^5{y}^4$

Step 1: Simplify and look for algebraic rewrites.

Here the equation is a simplified as you need it, and there are no algebra rewrites.

You always have to check though.

$x^3+y^2=x^5{y}^4$

Step 2: Wrap the problem in a set of parentheses and $\frac{d}{d\left(\mathit{whatever\; your\; respect\; to\; variable}\right)}$ the entire equation.

In this example, since you are being asked to find $\frac{\mathit{dy}}{dx}$, you will want to $\frac{d}{dx}$ the entire equation.

$\frac{d}{dx}(x^3+y^2=x^5{y}^4)$

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.

In this example you have 3-terms that we must take the derivative of.

$\frac{d}{\mathit{dx}}(x^3+y^2=x^5{y}^4)$

1) $x^3$: Power Rule

2) $y^2$: Power Rule

3) $x^5{y}^4$: Product Rule

Step 4: Start performing the individual derivative rules.

Keep this saying in mind as you go through the process, “Every time I take a derivative, I get extra stuff.”

1) You apply the power rule to $x^3$, and because you took a derivative, you get extra stuff. Since, you took the derivative of an x , your extra stuff is $\frac{d\overline{)x}}{dx}$.

2) You apply the power rule to $y^2$, and because you took a derivative, you get extra stuff. Since, you took the derivative of an y , your extra stuff is $\frac{d\overline{)y}}{dx}$.

3) Here you have x ’s and y ‘s being multiplied, which means we must apply the Product Rule: $fg‘+gf‘$.

In this situation it is only when we find the $f‘$ and the $g‘$, that we are taking a derivative and therefore get our extra stuff.

Bring all the pieces back together to create your derivative equation. This is not your final result, just the results of your derivatives of your bitesize pieces.

I have it broken up into 3 separate pieces in this example. Ideally you would want to get to where you can go immediately to the final form.

1) $x^3$: Power Rule

Derivative: ${3x}^2\frac{d\overline{)x}}{dx}$

2) $y^2$: Power Rule

Derivative: $2y\frac{d\overline{)y}}{dx}$

3) $x^5{y}^4$: Product Rule

$f=x^5$ $g=y^4$

$f‘={5x}^4\frac{d\overline{)x}}{dx}$ $g‘={4y}^3\frac{d\overline{)y}}{dx}$

Product Rule: $fg‘+gf‘$

$\left(x^5\right)\left({4y}^3\frac{d\overline{)y}}{dx}\right)+\left(y^4\right)\left({5x}^4\frac{d\overline{)x}}{dx}\right)$

$\frac{d}{\mathit{dx}}(x^3+y^2=x^5{y}^4)$

Derivative:

${3x}^2\frac{d\overline{)x}}{dx}+2y\frac{d\overline{)y}}{dx}=\left(x^5\right)\left({4y}^3\frac{d\overline{)y}}{dx}\right)+\left(y^4\right)\left({5x}^4\frac{d\overline{)x}}{dx}\right)$

${3x}^2\frac{\mathit{dx}}{\mathit{dx}}+2y\frac{\mathit{dy}}{\mathit{dx}}=\left(x^5\right)\left({4y}^3\frac{\mathit{dy}}{\mathit{dx}}\right)+\left(y^4\right)\left({5x}^4\frac{\mathit{dx}}{\mathit{dx}}\right)$

Step 5: Clean up the equation.

You work your way left to right through the equation, 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

Here we have several $\frac{\mathit{dx}}{\mathit{dx}}$ that we are able to cancel.

2) Rewrites

No rewrites.

3) Proper order

We can reorder the two terms on the right side of the equal, and also drop the extra parentheses.

${3x}^2\frac{\mathit{dx}}{\mathit{dx}}+2y\frac{\mathit{dy}}{\mathit{dx}}=\left(x^5\right)\left({4y}^3\frac{\mathit{dy}}{\mathit{dx}}\right)+\left(y^4\right)\left({5x}^4\frac{\mathit{dx}}{\mathit{dx}}\right)$

${3x}^2\frac{\overline{)\mathit{dx}}}{\overline{)\mathit{dx}}}+2y\frac{\mathit{dy}}{\mathit{dx}}=\left(x^5\right)\left({4y}^3\frac{\mathit{dy}}{\mathit{dx}}\right)+\left(y^4\right)\left({5x}^4\frac{\overline{)\mathit{dx}}}{\overline{)\mathit{dx}}}\right)$

${3x}^2+2y\frac{\mathit{dy}}{\mathit{dx}}=\left(x^5\right)\left({4y}^3\frac{\mathit{dy}}{\mathit{dx}}\right)+\left(y^4\right)\left({5x}^4\right)$

${3x}^2+2y\frac{\mathit{dy}}{\mathit{dx}}=\overline{)\left(x^5\right)\left({4y}^3\frac{\mathit{dy}}{\mathit{dx}}\right)}+\overline{)\left(y^4\right)\left({5x}^4\right)}$

${3x}^2+2y\frac{\mathit{dy}}{\mathit{dx}}=4x^5{y}^3\frac{\mathit{dy}}{\mathit{dx}}+{5x}^4{y}^4$

Step 6: Solve for your derivative piece.

In this example you are looking to find $\frac{\mathit{dy}}{\mathit{dx}}$. You need to get $\frac{\mathit{dy}}{\mathit{dx}}$ alone on one side of the equals.

i) Whenever you have multiples of what you need to solve for, i.e. multiple $\frac{\mathit{dy}}{\mathit{dx}}$’s, then the algebra move you want to perform is to get every term that has a $\frac{\mathit{dy}}{\mathit{dx}}$ on one side of the equals, and every term that does not have it on the other side of the equals. Break it up into pieces that have a $\frac{\mathit{dy}}{\mathit{dx}}$ and pieces that don’t have a $\frac{\mathit{dy}}{\mathit{dx}}$.

It doesn’t matter which side of the equal we move the pieces to. Here I will move the $4x^5{y}^3\frac{\mathit{dy}}{\mathit{dx}}$ to the left side and the ${3x}^2$ to the left side.

ii) Now that all the terms on the left side of the equals include a $\frac{\mathit{dy}}{\mathit{dx}}$, we can factor a $\frac{\mathit{dy}}{\mathit{dx}}$ out of those terms, and get down to a single $\frac{\mathit{dy}}{\mathit{dx}}.$

iii) Now that you have the $\frac{\mathit{dy}}{\mathit{dx}}$ on the outside of the parenthesis you can divide both sides by the piece inside the parenthesis to get $\frac{\mathit{dy}}{\mathit{dx}}$, alone on one side of the equals. Here you will divide both sides by $2y–4x^5{y}^3$, and that leaves $\frac{\mathit{dy}}{\mathit{dx}}$alone on the left side.

${3x}^2+2y\frac{\mathit{dy}}{\mathit{dx}}=4x^5{y}^3\frac{\mathit{dy}}{\mathit{dx}}+{5x}^4{y}^4$

$i)\begin{array}{c}{3x}^2\\ –{3x}^2\end{array}+\begin{array}{c}2y\frac{\mathit{dy}}{\mathit{dx}}\\ –4x^5{y}^3\frac{\mathit{dy}}{\mathit{dx}}\end{array}=\begin{array}{c}4x^5{y}^3\frac{\mathit{dy}}{\mathit{dx}}\\ –4x^5{y}^3\frac{\mathit{dy}}{\mathit{dx}}\end{array}+\begin{array}{c}{5x}^4{y}^4\\ –{3x}^2\end{array}$

$2y\frac{\mathit{dy}}{\mathit{dx}}–4x^5{y}^3\frac{\mathit{dy}}{\mathit{dx}}={5x}^4{y}^4–{3x}^2$

$\mathrm{ii})\mathrm{}2y\frac{\mathit{dy}}{\mathit{dx}}–4x^5{y}^3\frac{\mathit{dy}}{\mathit{dx}}={5x}^4{y}^4–{3x}^2$

$\frac{\mathit{dy}}{\mathit{dx}}(2y–4x^5{y}^3)={5x}^4{y}^4–{3x}^2$

$\mathit{iii}\left)\mathrm{}\frac{\mathit{dy}}{\mathit{dx}}\right(2y–4x^5{y}^3)={5x}^4{y}^4–{3x}^2$

$\frac{\mathit{dy}}{\mathit{dx}}\frac{\left(2y–4x^5{y}^3\right)}{2y–4x^5{y}^3}=\frac{{5x}^4{y}^4–{3x}^2}{2y–4x^5{y}^3}$

$\frac{\mathit{dy}}{\mathit{dx}}=\frac{{5x}^4{y}^4–{3x}^2}{2y–4x^5{y}^3}$

$\frac{\mathit{dy}}{\mathit{dx}}=\frac{{5x}^4{y}^4–{3x}^2}{2y–4x^5{y}^3}$

Final Result:

The derivative of $x^3+y^2=x^5{y}^4$ is $\frac{\mathit{dy}}{\mathit{dx}}=\frac{{5x}^4{y}^4–{3x}^2}{2y–4x^5{y}^3}$.

Meaning:

– The equation for finding the slope of any tangent line at any $\left(x,y\right)$ of $x^3+y^2=x^5{y}^4$ is $\frac{\mathit{dy}}{\mathit{dx}}=\frac{{5x}^4{y}^4–{3x}^2}{2y–4x^5{y}^3}$.

– The instantaneous rate of change for every $\left(x,y\right)$ of $x^3+y^2=x^5{y}^4$, is found by using the derivative equation,

$\frac{\mathit{dy}}{\mathit{dx}}=\frac{{5x}^4{y}^4–{3x}^2}{2y–4x^5{y}^3}$.

– The slope of $x^3+y^2=x^5{y}^4$at any single $\left(x,y\right)$ can be found by plugging it into the derivative, $\frac{\mathit{dy}}{\mathit{dx}}=\frac{{5x}^4{y}^4–{3x}^2}{2y–4x^5{y}^3}$.