**Q: F(x) = (3x²+1) ^{3}/(x^{2}-1)^{4}. Find the derivative of F(x), F'(x).**

A: The most important thing to do is to identify the “main rule” that we are going to use. Since I see a fraction, I am going to use the quotient rule.

**The quotient rule tells us that if you have f/g, that the derivative is (gf’ – g’f)/g ^{2}**

So, set up the following table:

f = (3x²+1)^{3} g = (x^{2}-1)^{4}

f’ = g’ =

Now we need to take the derivatives of f and g separately to find f’ and g’. Taking these derivatives will require the chain rule. Some teachers teach the chain rule by substituting a “u”, and various other methods. I don’t like that way. I think of the chain rule as an “onion rule”… You need to peel off layers. Take the derivative of the outside, times the derivative of the inside!

Let me try this in words as well before I do it: If f = (something)^{3}, then f’ = 4(something)^{3}*(derivative of that something)

So,

f = (3x²+1)^{3}

f’ = 3(3x²+1)^{2}(6x) = 18x(3x²+1)^{2}

g = (x^{2}-1)^{4}

g’ = 4(x^{2}-1)^{3}(2x) = 8x(x^{2}-1)^{3}

So, our original function looked like f/g… The derivative of this, as said before, is:

(gf’ – g’f) / g^{2} = [(x^{2}-1)^{4}*18x(3x²+1)^{2} – 8x(x^{2}-1)^{3}(3x²+1)^{3}] / (((x^{2}-1)^{4})^{2}

= [18x*(x^{2}-1)^{4}(3x²+1)^{2} – 8x(x^{2}-1)^{3}(3x²+1)^{3}] / (x^{2}-1)^{8}