Need help with derivatives

**jkh1919** · October 6th 2012, 11:39 AM

Hello everyone, I have a few derivatives problems I could use some guidance with. I know the procedures for finding derivatives (product/quotient rule, etc...), but for some reason I feel like I am doing things the hard way.

Here is the problem I am stuck on now, but I am studying for a huge test on Monday and I'm sure I will be asking quite a bit of questions between now and then.

Find dy/dx at x=1
(x+2)(x^2+1)(x^3+2)

Long story short, I used the product rule to find the derivative of the 3 parts, and then plugged in 1 for x. My answer was 50 and it is incorrect. Can someone maybe show me the correct steps to work this problem? The more detail the better, because I need to see everything about how a problem works. I used the product rule to find the derivatives of the first 2 parts, then took that and used the power rule again. Obviously this is not correct, and now I am wondering just how you are supposed to solve this problem. Do I need to first multiply 2 of the 3 terms together and then use the power rule?

Help is appreciated.

**MaxJasper** · October 6th 2012, 11:53 AM

$f'(x)=3 x^2 (2+x) \left(1+x^2\right)+2 x (2+x) \left(2+x^3\right)+\left(1+x^2\right) \left(2+x^3\right)=2+8 x+12 x^2+4 x^3+10 x^4+6 x^5$

**TheEmptySet** · October 6th 2012, 11:57 AM

Originally Posted by jkh1919

Hello everyone, I have a few derivatives problems I could use some guidance with. I know the procedures for finding derivatives (product/quotient rule, etc...), but for some reason I feel like I am doing things the hard way.

Here is the problem I am stuck on now, but I am studying for a huge test on Monday and I'm sure I will be asking quite a bit of questions between now and then.

Find dy/dx at x=1
(x+2)(x^+1)(x^3+2)

Long story short, I used the product rule to find the derivative of the 3 parts, and then plugged in 1 for x. My answer was 50 and it is incorrect. Can someone maybe show me the correct steps to work this problem? The more detail the better, because I need to see everything about how a problem works. I used the product rule to find the derivatives of the first 2 parts, then took that and used the power rule again. Obviously this is not correct, and now I am wondering just how you are supposed to solve this problem. Do I need to first multiply 2 of the 3 terms together and then use the power rule?

Help is appreciated.

Are you familiar with logarithmic derivatives?

$y=(x+2)(x^2+1)(x^3+2)$

If we take the natural log of both sides we get

$\ln(y)=\ln(x+2)+\ln(x^2+1)+\ln(x^3+2)$

Now if we take the derivative of both sides we get

$\frac{y'}{y}=\frac{1}{x+2}+\frac{2x}{x^2+1}+\frac{ 3x^2}{x^3+2}$

This gives

$y'=y \left( \frac{1}{x+2}+\frac{2x}{x^2+1}+\frac{3x^2}{x^3+2} \right)$

If we evaluate this at zero we get

$y'(1)=y(1) \left( \frac{1}{1+2}+\frac{2(1)}{1^2+1}+\frac{3(1)^2}{1^3 +2} \right)$

$y'(1)=(3)(2)(3)\left( \frac{1}{3}+\frac{2}{2}+\frac{3}{3}\right)=42$

**jkh1919** · October 6th 2012, 12:10 PM

Originally Posted by TheEmptySet

Are you familiar with logarithmic derivatives?

$y=(x+2)(x^2+1)(x^3+2)$

If we take the natural log of both sides we get

$\ln(y)=\ln(x+2)+\ln(x^2+1)+\ln(x^3+2)$

Now if we take the derivative of both sides we get

$\frac{y'}{y}=\frac{1}{x+2}+\frac{2x}{x^2+1}+\frac{ 3x^2}{x^3+2}$

This gives

$y'=y \left( \frac{1}{x+2}+\frac{2x}{x^2+1}+\frac{3x^2}{x^3+2} \right)$

If we evaluate this at zero we get

$y'(1)=y(1) \left( \frac{1}{1+2}+\frac{2(1)}{1^2+1}+\frac{3(1)^2}{1^3 +2} \right)$

$y'(1)=(3)(2)(3)\left( \frac{1}{3}+\frac{2}{2}+\frac{3}{3}\right)=42$

Thanks. I am not familiar with logarithmic derivatives, and actually that is the first time I have heard that term. I appreciate the help using this example, and will definitely note it for future reference, but right now I am really trying to perfect the use of the chain, product, and quotient rules, and need someone to explain to me the steps that were taken to solve the problems.

Here is the response above:

Originally Posted by MaxJasper

$f'(x)=3 x^2 (2+x) \left(1+x^2\right)+2 x (2+x) \left(1+x^3\right)+\left(1+x^2\right) \left(1+x^3\right)=1+4 x+9 x^2+4 x^3+10 x^4+6 x^5$

This is the product rule, correct? I can see what was done to get the answer, but some things I am not so clear about.

In particular, this:

+2 x (2+x) (1+x^3)+(1+x^2) (1+x^3)

Can someone point out to me why the term (1+x^3) is there?

I can see that the steps used to solve this problem were to take the derivative of the 3rd term, multiply it by the first two, then ADD it to, the derivative of the 2nd term, multiplied by the other two terms, but I can't tell what happened after that. I would think that we'd still need to take the derivative of the 1st term, multiply it by the other 2 terms, and add that to the results from the other two. So pretty much I understand, up until this point:

3 x^2 (2+x)(1+x^2)+2 x (2+x) (1+x^3)

But still can not figure out where the (1+x^3) came from

**alane1994** · October 8th 2012, 09:46 AM

I went through and multiplied and combined from the original equation. x⁶+2x⁵+x⁴+4x³+2x²+2x+4(no clue if this is how you do it)...
Then I took the derivative of that. 6x⁵+10x⁴+4x³+12x²+4x+2
From the derivative of that, I plugged 1 into it for x. From that, I got an answer of 38... I hope that made sense...

**MaxJasper** · October 8th 2012, 10:18 AM

Your method of first expanding then taking derivative: if such a problem appears in your exam then all your time is wasted on answering just one question! Get efficient. f'(1)=42 and 38 is wrong because your expansion is wrong after all the efforts. Such errors usually occur when expanding these kinds of polynomials.

**HallsofIvy** · October 8th 2012, 11:33 AM

I think the simplest way to do this is to use the "extended" product rule: (fgh)'= f'gh+ fg'h+ fgh'
(x(x^2+ 1)(x^3+ 2))'= (1)(x^2+ 1)(x^3+ 2)+ x(2x)(x^3+ 2)+ x(x^2+ 1)(3x^2)

Now set x= 1 in that.

**alane1994** · October 8th 2012, 11:35 AM

I have never heard of the extended product rule... I will have to ask my professor about that! This seems like a much easier way to do this, seeing as natural logs aggravate me to no end...

**hollywood** · October 8th 2012, 04:41 PM

The extended product rule is just the regular product rule applied twice:
$(fgh)'$
$=((fg)(h))'$
$=(fg)'h+fgh'$
$=(f'g+fg')h+fgh'$
$=f'gh+fg'h+fgh'$

- Hollywood

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