Stumped Calc Students: Can You Solve This Diff. Equation?

[Aresius]

This is a problem that has stumped my entire class of Calc 1 students and two Calc 2 students.

$$Find \frac {dy} {dx}$$

$$y = \frac {(2x+3)^3} {(4x^2-1)^8}$$

I know that the answer is (from the textbook, but I don't know how it got there)

$$-\frac {2(2x+3)^2(52x^2+96x+3)} {(4x^2-1)^9}$$

I'll attempt to show my work in the next post.

 

[Aresius]

First I took the derivative using the chain rule of both the numerator and the denominator and applied them into the quotient rule.

$$f' (2x+3)^3 = 6(2x+3)^2$$
and
$$g' (4x^2-1)^8 = 256x(4x-1)^7$$

Applying the quotient rule I get this

$$\frac {6(4x^2-1)^8(2x+3)^2 - 256x(4x-1)^7(2x+3)^3} {[(4x^2-1)^8]^2}$$

I have no clue where they went from there...

 

[arildno]

g' is wrong, your coefficient in front of the seventh' power parenthesis should be 64, not 256.

 

[Mulder]

There should be a "squared" in the (4x - 1) bracket on the right of the top line of the quotient, then it just cancels.

 

[BobG]

First I took the derivative using the chain rule of both the numerator and the denominator and applied them into the quotient rule.

$$f' (2x+3)^3 = 6(2x+3)^2$$
and
$$g' (4x^2-1)^8 = 256x(4x-1)^7$$

Applying the quotient rule I get this

$$\frac {6(4x^2-1)^8(2x+3)^2 - 256x(4x-1)^7(2x+3)^3} {[(4x^2-1)^8]^2}$$

I have no clue where they went from there...

As arildno said, your g' is wrong. Your denominator is $$(4x^2-1)^{16}$$
You can divide all the terms by a $$(4x^2-1)^7$$. After that, it's just some factoring and multiplication in the numerator.

 

[Aresius]

I thought that because of the chain rule you have to take the derivative of the derivative of 4x^2 - 1 and multiply it by the others in sequence.

God I hate the chain rule...

I think I messed up g' more than I thought... You cannot touch the damn original function!

 

[TD]

Did you get there now or still need help?

 

[Aresius]

Wait, I already did that, never mind...

 

[Aresius]

Ok am I on the right track here?

$$\frac {6(4x^2-1)(2x+3)^2 - 64x(2x+3)^3} {(4x^2-1)^9}$$

 

[Aresius]

You must take the derivative of 8x as per the chain rule and then multiply it into that term.

Am I right?

In that case I would have a 512 in front of that x, bleh

 

[TD]

Yes, seems to be correct so far

 

[Aresius]

Was that to my last post? Because I somehow doubt it :(

 

[TD]

It was in referrence to

Ok am I on the right track here?

$$\frac {6(4x^2-1)(2x+3)^2 - 64x(2x+3)^3} {(4x^2-1)^9}$$

So you're doing ok so far.

 

[Aresius]

But that I don't get, as per the chain rule you must first get 8 from g', then 8x from what was in the brackets, then another 8 from 8x. Multiply them together and you get 512x, why is it still 64x?

 

[TD]

You have a bit too many 8's then ...

$$\left( {\frac{{\left( {2x + 3} \right)^3 }} {{\left( {4x^2 - 1} \right)^8 }}} \right)^\prime = \frac{{6\left( {2x + 3} \right)^2 \left( {4x^2 - 1} \right)^8 - 64x\left( {4x^2 - 1} \right)^7 \left( {2x + 3} \right)^3 }} {{\left( {4x^2 - 1} \right)^{16} }} = \frac{{6\left( {2x + 3} \right)^2 \left( {4x^2 - 1} \right) - 64x\left( {2x + 3} \right)^3 }} {{\left( {4x^2 - 1} \right)^9 }}$$

One comes from the power (x^8 => 8x^7) and the other one from (4x²)' = 8x, which gives 8*8x = 64x.

 

[Aresius]

Yes, and then you must take the derivative of 8x and multiply that on because of the chain rule, right?

I thought that as long as you can take a rational derivative with the chain rule, you should.

 

[Aresius]

$$8(4x^2-1)^7(8x)(8)$$ is how we do chain rule in class.

Perhaps I'm horribly mistaken and I've forgotten part of the rules of the chain rule...

 

[TD]

We're finished at that point, no more 8's to multiply...

$$\left( {\left( {4x^2 - 1} \right)^8 } \right)^\prime = 8\left( {4x^2 - 1} \right)^7 \cdot \left( {4x^2 - 1} \right)^\prime = 8\left( {4x^2 - 1} \right)^7 \cdot 8x = 64x\left( {4x^2 - 1} \right)^7$$

Ah no, you have an 8 extra. The derivative of 4x² is just 8x; not 8x*8 since (4x²)' = 4*(x²)' = 4*2x = 8x.

 

[Aresius]

Am I right so far in simplifying (using pascal's triangle and binomial theorem)

$$\frac {6(4x^2-1)(2x+3)^2 - 64x(8x^3+36x^2+54x+27)} {4x^2-1)^9}$$

 

[TD]

Yes, that is correct. You also could've factored out $(2x+3)^2$ first.

 

[Aresius]

The answer to the problem still has the (2x+3)^2 so I left it as it is, am I wrong?

 

[TD]

It's not wrong but it'll be easier to factor it out. As you say, the factor is still there in the answer.

$$\frac{{6\left( {2x + 3} \right)^2 \left( {4x^2 - 1} \right) - 64x\left( {2x + 3} \right)^3 }} {{\left( {4x^2 - 1} \right)^9 }} = \frac{{2\left( {2x + 3} \right)^2 \left( {3\left( {4x^2 - 1} \right) - 32x\left( {2x + 3} \right)} \right)}} {{\left( {4x^2 - 1} \right)^9 }}$$

Now work out everything in those last parenthesis.

 

[Aresius]

Got it!

Thanks!

 

[TD]

Great

 

[BobG]

It's sometimes easier to use substitution to organize yourself.

I know you don't think the derivative of $$4x^2-1$$ is 8*8x. Your problem was the same a lot of people have with the chain rule - once they get going, they don't know when to stop.

If you had broken it out as:

$$4x^2-1 = u$$ then you have two problems:

You need to find the derivative of $$u^8$$

You need to find the derivative of $$4x^2-1$$

 

[Aresius]

I don't know why I thought this, but I thought as long as you had a term that could be counted as a function inside the original function, you should take it's derivative and multiply in series as per chain rule.

I think I see what you mean though, you only take derivative of what has not been derived already. The original function never changes.