Example 1: Quotient Rule Derivative

Differentiate $f\left(x\right)=\frac{{(3x}^3–6x+9)}{(5x^4+3x^2)}$ with respect to x.

Step 1: Simplify and look for algebraic rewrites.

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

You always have to check though.

$f\left(x\right)=\frac{{(3x}^3–6x+9)}{(5x^4+3x^2)}$

Step 2: Identify your primary rule as the quotient rule.

Looking at this problem, we can see that it fits the general shape of a quotient rule. We have $\frac{\left(\mathit{SOMETHING}\right)}{\left(\mathit{SOMETHING}\right)}$.

$f\left(x\right)=\frac{\left(\mathit{SOMETHING}\right)}{\left(\mathit{SOMETHING}\right)}$

$f\left(x\right)=\frac{{(3x}^3–6x+9)}{(5x^4+3x^2)}$

Step 3: Label your top SOMETHING hi , then label the bottom SOMETHING low .

$f\left(x\right)=\frac{{\overline{){(3x}^3–6x+9)}}^{\mathit{hi}}}{{\overline{)(5x^4+3x^2)}}^{\mathit{low}}}$

Step 4: Break the problem up into two these two bite-size problems.

$\mathit{hi}=\mathit{SOMETHING}$ and $\mathit{low}=\mathit{SOMETHING}$

$\mathit{hi}={3x}^3–6x+9$                $\mathit{low}=5x^4+3x^2$

Step 5: Take the derivatives of those two bite-size problems to find your dhi and your dlow .

The derivatives of hi and low only require us to apply the power rule.

Bring the power down.

Subtract 1 from the power.

$\mathit{hi}={3x}^3–6x+9$                $\mathit{low}=5x^4+3x^2$

$\mathit{dhi}={9x}^2–6$                        $\mathit{dlow}=20x^3+6x$

Step 6: Bring it all back together following the quotient rulerecipe: $\frac{\left(\mathit{low}\right)\left(\mathit{dhi}\right)\mathit{–}\left(\mathit{hi}\right)\left(\mathit{dlow}\right)}{{\left(\mathit{low}\right)}^2}$.

$f\prime \left(x\right)=\mathrm{ }\frac{\left(\mathit{low}\right)\left(\mathit{dhi}\right)\mathit{–}\left(\mathit{hi}\right)\left(\mathit{dlow}\right)}{{\left(\mathit{low}\right)}^2}\mathrm{ }$

$f\prime \left(x\right)=\mathrm{ }\frac{(5x^4+3x^2)({9x}^2–6)\mathit{–}({3x}^3–6x+9)(20x^3+6x)}{{(5x^4+3x^2)}^2}$

Final Result:

The derivative of $f\left(x\right)=\frac{{(3x}^3–6x+9)}{(5x^4+3x^2)}$  is $f\prime \left(x\right)=\mathrm{ }\frac{(5x^4+3x^2)({9x}^2–6)\mathit{–}({3x}^3–6x+9)(20x^3+6x)}{{(5x^4+3x^2)}^2}$.

Most of the time you will want to leave the derivative as it is and not try to expand it or simplify it.

My rule of thumb is if someone wants the derivative give them this unless they ask you to simplify further.

If you need to do more work, like finding a second derivative, you would want to consider expanding and simplifying to make your life easier.

Meaning:

–          The equation for finding the slope of any tangent line at any x-value of $f\left(x\right)=\frac{{(3x}^3–6x+9)}{(5x^4+3x^2)}$  is $f\prime \left(x\right)=\mathrm{ }\frac{(5x^4+3x^2)({9x}^2–6)\mathit{–}({3x}^3–6x+9)(20x^3+6x)}{{(5x^4+3x^2)}^2}$.

–          The instantaneous rate of change for every x-value of $f\left(x\right)=\frac{{(3x}^3–6x+9)}{(5x^4+3x^2)}$, is found by using the derivative equation, $f\prime \left(x\right)=\mathrm{ }\frac{(5x^4+3x^2)({9x}^2–6)\mathit{–}({3x}^3–6x+9)(20x^3+6x)}{{(5x^4+3x^2)}^2}$.

–          The slope of $f\left(x\right)=\frac{{(3x}^3–6x+9)}{(5x^4+3x^2)}$at any single x-value can be found by plugging it into the derivative, $f\prime \left(x\right)=\mathrm{ }\frac{(5x^4+3x^2)({9x}^2–6)\mathit{–}({3x}^3–6x+9)(20x^3+6x)}{{(5x^4+3x^2)}^2}$.