Example 3: Quotient-Chain

Differentiate $f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$ 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.

$f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$

Step 2: Identify your primary derivative rule.

Squint your eyes and decided what “shape” this problem most has. Something divided by Something.

$y=\frac{\mathit{\left(}\mathit{ }\mathit{\right)}}{\mathit{\left(}\mathit{ }\mathit{\right)}}$

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

We can see that there are some individual pieces that look like chain rules, but that is not the primary (overall) action occurring.

This tells us our primary rule is the quotient rule.

So, we follow the quotient rule process.

$f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$

$\frac{\mathit{\left(}\mathit{ }\mathit{\right)}}{\mathit{\left(}\mathit{ }\mathit{\right)}}$

Step 3: Break the problem up into its bite-size chunks based on the primary derivative rule you chose in Step 2.

Since we are applying the quotient rule, we will break apart our equation, labeling the top something hi and label the bottom something low .

This means we will now need to do what is necessary to find the derivatives of those bite-size chunks in order to get the dhi and dlow we need for our quotient rule recipe.

Quotient Rule:   $\frac{\left(\mathit{low}\right)\left(\mathit{dhi}\right)\mathit{–}\left(\mathit{hi}\right)\left(\mathit{dlow}\right)}{{\left(\mathit{low}\right)}^2}$

$f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$

$f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$

$f\left(x\right)=\frac{{\overline{){\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}}^{\mathit{hi}}}{{\overline{){\mathrm{(}5x^4+3x^2\mathrm{)}}^7}}^{\mathit{low}}}$

$\mathit{hi}={\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6$

$\mathit{dhi}=\overline{)}$

$\mathit{low}={\mathrm{(}5x^4+3x^2\mathrm{)}}^7$

$\mathit{dlow}=\overline{)}$

Step 4: Identify and apply any secondary derivative rule you need to find the derivative of each of your bite-size chunks.

We now treat the f and g equations like two brand new math problems and address how to take those two derivatives by identifying their derivative rule.

In this situation both the f and g equations fit the chain rule shape.

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

So, we follow the chain rule process.

Chain Rule: ( DIN )( DOUT )

Step 1: Break them up into two bite-size problems. The IN piece and the OUT piece.

Step 2: Take the derivatives of those two pieces to find your DIN (derivative of the inside piece)and your DOUT (derivative of the outside piece).

Step 3: Bring it all back together following the Chain Rule: ( DIN )( DOUT )

$\mathit{hi}={\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6$

$\mathit{IN}={7x}^3–45x^2+35x$

$\mathit{OUT}={({7x}^3–45x^2+35x)}^6$

$\mathit{DIN}={21x}^2–90x+35$

  $\mathit{DOUT}=6{({7x}^3–45x^2+35x)}^{\mathrm{}5}$

$\mathit{dhi}=\mathrm{(}{21x}^2–90x+35\mathrm{)}\mathrm{(}6{({7x}^3–45x^2+35x)}^{\mathrm{}5}\mathrm{)}$

$\mathit{low}={\mathrm{(}5x^4+3x^2\mathrm{)}}^7$

$\mathit{IN}=5x^4+3x^2$

$\mathit{OUT}={(5x^4+3x^2)}^7$

$\mathit{DIN}=20x^3+6x$

$\mathit{DOUT}={7(5x^4+3x^2)}^{\mathrm{}6}$

$\mathit{dlow}=\mathrm{(}20x^3+6x\mathrm{)}\mathrm{(}{7(5x^4+3x^2)}^{\mathrm{}6}\mathrm{)}$

Step 5: Bring it all back together following the primary derivative rule.

Remember you were really in the middle of a product rule when this all started, and you now just found your f’ and g’ that you needed.

Now that we have it all we put it together following the Product Rule: $f{g}^{\prime }+g{f}^{\prime }$

$f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$

$f\left(x\right)=\frac{{\overline{){\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}}^{\mathit{hi}}}{{\overline{){\mathrm{(}5x^4+3x^2\mathrm{)}}^7}}^{\mathit{low}}}$

$\mathit{hi}={\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6$   $\mathit{low}={\mathrm{(}5x^4+3x^2\mathrm{)}}^7$

$\mathit{dhi}=\mathrm{(}{21x}^2–90x+35\mathrm{)}\mathrm{(}6{({7x}^3–45x^2+35x)}^{\mathrm{}5}\mathrm{)}$   $\mathit{dlow}=\mathrm{(}20x^3+6x\mathrm{)}\mathrm{(}{7(5x^4+3x^2)}^{\mathrm{}6}\mathrm{)}$

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

$f\prime \left(x\right)=\frac{{\overline{){\mathrm{(}5x^4+3x^2\mathrm{)}}^7}}^{\mathit{low}}{\overline{)\mathrm{(}{21x}^2–90x+35\mathrm{)}\bullet 6{({7x}^3–45x^2+35x)}^{\mathrm{}5}}}^{\mathit{dhi}}\mathit{–}{\overline{){\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}}^{\mathit{hi}}{\overline{)\mathrm{(}20x^3+6x\mathrm{)}\bullet {7(5x^4+3x^2)}^{\mathrm{}6}}}^{\mathit{dlow}}}{{\left({\overline{){\mathrm{(}5x^4+3x^2\mathrm{)}}^7}}^{\mathit{low}}\right)}^2}$

$f^{\prime }\mathrm{(}x\mathrm{)}=\frac{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7\bullet \mathrm{(}{21x}^2–90x+35\mathrm{)}\bullet 6{({7x}^3–45x^2+35x)}^{\mathrm{}5}\mathit{–}{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6\bullet \mathrm{(}20x^3+6x\mathrm{)}\bullet {7(5x^4+3x^2)}^{\mathrm{}6}}{{\left({\mathrm{(}5x^4+3x^2\mathrm{)}}^7\right)}^2}$

 Final Result:

The derivative of $f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$  is $f\prime \left(x\right)= \frac{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7\bullet \mathrm{(}{21x}^2–90x+35\mathrm{)}\bullet 6{({7x}^3–45x^2+35x)}^{\mathrm{}5}\mathit{–}{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6\bullet \mathrm{(}20x^3+6x\mathrm{)}\bullet {7(5x^4+3x^2)}^{\mathrm{}6}}{{\left({\mathrm{(}5x^4+3x^2\mathrm{)}}^7\right)}^2}$.

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 $f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$  is $f\prime \left(x\right)= \frac{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7\bullet \mathrm{)}{21x}^2–90x+35\mathrm{)}\bullet 6{({7x}^3–45x^2+35x)}^{\mathrm{}5}\mathit{–}{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6\bullet \mathrm{(}20x^3+6x\mathrm{)}\bullet {7(5x^4+3x^2)}^{\mathrm{}6}}{{\left({\mathrm{(}5x^4+3x^2\mathrm{)}}^7\right)}^2}$.

–          The instantaneous rate of change for every x-value of $f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$, is found by using the derivative equation, $f\prime \left(x\right)= \frac{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7\bullet \mathrm{(}{21x}^2–90x+35\mathrm{)}\bullet 6{({7x}^3–45x^2+35x)}^{\mathrm{}5}\mathit{–}{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6\bullet \mathrm{(}20x^3+6x\mathrm{)}\bullet {7(5x^4+3x^2)}^{\mathrm{}6}}{{\left({\mathrm{(}5x^4+3x^2\mathrm{)}}^7\right)}^2}$.

–          The slope of $f\left(x\right)=\frac{{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6}{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7}$at any single x-value can be found by plugging it into the derivative, $f\prime \left(x\right)= \frac{{\mathrm{(}5x^4+3x^2\mathrm{)}}^7\bullet \mathrm{(}{21x}^2–90x+35\mathrm{)}\bullet 6{({7x}^3–45x^2+35x)}^{\mathrm{}5}\mathit{–}{\mathrm{(}{7x}^3–45x^2+35x\mathrm{)}}^6\bullet \mathrm{(}20x^3+6x\mathrm{)}\bullet {7(5x^4+3x^2)}^{\mathrm{}6}}{{\left({\mathrm{(}5x^4+3x^2\mathrm{)}}^7\right)}^2}$.