Example 4: Chain-Quotient

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

Step 1: Simplify and look for algebraic rewrites.

Anytime we see a root in a problem we know that we need to use the root rewrite before we do anything else.

$\sqrt[\mathit{root}]{x^{\mathit{power}}}=x^{\frac{\mathit{power}}{\mathit{root}}}$

$\sqrt{x}$

Remember that there is an unwritten power and root in this situation.

$\sqrt[2]{x^1}=x^{\frac{1}{2}}$

Since the entire fraction is inside the square root, we wrap the whole thing in a set of parentheses, and raise it to the $\frac{1}{2}$ power.

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

$f\left(x\right)=\sqrt[\mathit{2}]{{\left(\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\right)}^{\mathit{1}}}$

$f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{\mathit{1}}{\mathit{2}}}$

Step 2: Identify your primary derivative rule.

Squint your eyes and decided what “shape” this problem most has. Here we primarily have one large parentheses with something inside it, and it is all raised to a power.

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

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

This tells us our primary rule is the chain rule.

So, we follow the chain rule process.

$f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

$f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

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

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 chain rule, we will break apart our equation, labeling one of the somethings IN and label the other something OUT .

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 ( DIN ) and ( DOUT ) we need for our chain rule recipe.

Chain Rule: (DIN) (DOUT)

$f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

$f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

$\mathit{IN}=\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}$

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

$\mathit{OUT}={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

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

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 IN and OUT equations like two brand new math problems and address how to take those two derivatives by identifying their derivative rule.

Here the IN equation will require the quotient rule and the OUT equation fits the power rule.

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

So, we follow the product rule process.

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

Step 1: Label the top something hi and label the bottom something low

Step 2: Break them up into two bite-size problems.

Step 3: Take the derivatives of those two pieces to find your dhi and dlow .

Step 4: Bring it all back together following the Quotient Rule:   $\frac{\left(\mathit{low}\right)\left(\mathit{dhi}\right)\mathit{–}\left(\mathit{hi}\right)\left(\mathit{dlow}\right)}{{\left(\mathit{low}\right)}^2}$

We then follow the power rule process to find the DOUT .

Bring the power down.

Subtract 1 from the power.

$\mathit{IN}=\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}$

$\mathit{IN}=\frac{{\overline{){7x}^3–45x^2+35x}}^{\mathit{hi}}}{{\overline{)5x^4+3x^2}}^{\mathit{low}}}$

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

$\mathit{dhi}={21x}^2–90x+35$        $\mathit{dlow}=20x^3+6x$

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

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

$\mathit{OUT}={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

$\mathit{DOUT}=\frac{1}{2}{\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{–\frac{1}{2}}$

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

Remember you were really in the middle of a chani rule when this all started, and you now just found your DIN and DOUT that you needed.

Now that we have it all we put it together following the Chain Rule: (DIN) (DOUT)

$f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

$f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

$\mathit{IN}=\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}$

$\mathit{OUT}={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$

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

$\mathit{DOUT}=\frac{1}{2}{\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{–\frac{1}{2}}$

$f\prime \left(x\right)=\mathit{(DIN)}\mathit{(DOUT)}$

$f\prime \left(x\right)= {\overline{)\frac{(5x^4+3x^2)({21x}^2–90x+35)\mathit{–}({7x}^3–45x^2+35x)(20x^3+6x)}{{(5x^4+3x^2)}^2}\mathrm{ }}}^{\mathit{DIN}}{\overline{)\frac{1}{2}{\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{–\frac{1}{2}}}}^{\mathit{DOUT}}$

$f^{\prime }\mathrm{(}x\mathrm{)}=\frac{(5x^4+3x^2)({21x}^2–90x+35)\mathit{–}({7x}^3–45x^2+35x)(20x^3+6x)}{{(5x^4+3x^2)}^2}\bullet \frac{1}{2}{\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{–\frac{1}{2}}$

 Final Result:

The derivative of $f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$  is $f\prime \left(x\right)=\frac{(5x^4+3x^2)({21x}^2–90x+35)\mathit{–}({7x}^3–45x^2+35x)(20x^3+6x)}{{(5x^4+3x^2)}^2}\bullet \frac{1}{2}{\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{–\frac{1}{2}}$.

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 $f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$  is $f\prime \left(x\right)=\frac{(5x^4+3x^2)({21x}^2–90x+35)\mathit{–}({7x}^3–45x^2+35x)(20x^3+6x)}{{(5x^4+3x^2)}^2}\bullet \frac{1}{2}{\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{–\frac{1}{2}}$.

–          The instantaneous rate of change for every x-value of $f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$, is found by using the derivative equation, $f\prime \left(x\right)=\frac{(5x^4+3x^2)({21x}^2–90x+35)\mathit{–}({7x}^3–45x^2+35x)(20x^3+6x)}{{(5x^4+3x^2)}^2}\bullet \frac{1}{2}{\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{–\frac{1}{2}}$.

–          The slope of $f\left(x\right)={\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{\frac{1}{2}}$at any single x-value can be found by plugging it into the derivative, $f\prime \left(x\right)=\frac{(5x^4+3x^2)({21x}^2–90x+35)\mathit{–}({7x}^3–45x^2+35x)(20x^3+6x)}{{(5x^4+3x^2)}^2}\bullet \frac{1}{2}{\mathrm{(}\frac{{7x}^3–45x^2+35x}{5x^4+3x^2}\mathrm{)}}^{–\frac{1}{2}}$.