1. ## The Chain Rule

Hey.. I'm having some serious problems with the Chain Rule! It's my 2nd day on it and I'm getting a little frustrated on these extra problems I am doing but can't get right...

1. x^2√(9-x^2)
I know this is the product rule but I don't really get it.. Whenever I use the product rule then take the derivative of √(9-x^2) I just get LOST!!!

2. secx^2

I always get the wrong answer everytime and can't figure it out..

h'(x)= secxtanx X 2(secx)
= 2sec^2xtanx

The answer shows up as 2xsecx^2tanx^2

help!! I need help understanding the chain rule and I got no idea how to do problems like these!!

2. Originally Posted by elpermic
Hey.. I'm having some serious problems with the Chain Rule! It's my 2nd day on it and I'm getting a little frustrated on these extra problems I am doing but can't get right...

1. x^2√(9-x^2)
I know this is the product rule but I don't really get it.. Whenever I use the product rule then take the derivative of √(9-x^2) I just get LOST!!!
If you're new to this, I would approach it this way.

We have $\displaystyle f(x)=x^2\sqrt{9-x^2}$

Let $\displaystyle u(x)=x^2$ and $\displaystyle v(x)=\sqrt{9-x^2}$

We see that $\displaystyle u'(x)=2x$ and to find $\displaystyle v'(x)$, apply the chain rule.

Note that the chain rule is $\displaystyle \frac{d}{\,dx}\left[f(g(x))\right]=f'(g(x))\cdot g'(x)$.

So we see that $\displaystyle v'(x)=\tfrac{1}{2}(9-x^2)^{-\frac{1}{2}}\cdot\frac{d}{\,dx}\left[9-x^2\right]\implies v'(x)=\frac{1}{2\sqrt{9-x^2}}\cdot(-2x)=-\frac{2x}{2\sqrt{9-x^2}}$ $\displaystyle =-\frac{x}{\sqrt{9-x^2}}$

Now substitute $\displaystyle u(x),~v(x),~u'(x),~\text{and}~v'(x)$ into the equation for the product rule:

If $\displaystyle f(x)=u(x)v(x)$, then $\displaystyle f'(x)=u(x)v'(x)+v(x)u'(x)$

Can you take it from here?

2. secx^2

I always get the wrong answer everytime and can't figure it out..

h'(x)= secxtanx X 2(secx)
= 2sec^2xtanx

The answer shows up as 2xsecx^2tanx^2

help!! I need help understanding the chain rule and I got no idea how to do problems like these!!
I don't see anything wrong with your answer. Make sure you copied down the problem correctly...otherwise, I'd have to say the book's answer is wrong.

--Chris

3. For the first one put $\displaystyle f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6},$ hence $\displaystyle f'(x)=\frac{1}{2\sqrt{9x^{4}-x^{6}}}\cdot \left( 9x^{4}-x^{6} \right)'=\frac{9\cdot 4x^{3}-6x^{5}}{2x^{2}\sqrt{9-x^{2}}}=\frac{18x-3x^{3}}{\sqrt{9-x^{2}}}.$

4. Originally Posted by Krizalid
For the first one put $\displaystyle f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6},$ hence $\displaystyle f'(x)=\frac{1}{2\sqrt{9x^{4}-x^{6}}}\cdot \left( 9x^{4}-x^{6} \right)'=\frac{9\cdot 4x^{3}-6x^{5}}{2x^{2}\sqrt{9-x^{2}}}=\frac{18x-3x^{3}}{\sqrt{9-x^{2}}}.$
Ha! I was thinking about that...but he was working with product and chain rules...so I did it the way he was doing it. It is always good to see another approach

--Chris

5. I was working with your way and I still don't get how to do it...

f'(x)= (2x)(√(9-x^2) + (-x/2√(9-x^2))(x^2)
= 2x√(9-x^2) - (-x^3/2√(9-x^2))

Did I do something wrong??

6. Originally Posted by Krizalid
For the first one put $\displaystyle f(x)=x^2\sqrt{9-x^2}=\sqrt{9x^4-x^6},$ hence $\displaystyle f'(x)=\frac{1}{2\sqrt{9x^{4}-x^{6}}}\cdot \left( 9x^{4}-x^{6} \right)'=\frac{9\cdot 4x^{3}-6x^{5}}{2x^{2}\sqrt{9-x^{2}}}=\frac{18x-3x^{3}}{\sqrt{9-x^{2}}}.$
That is the answer but how do you get it using the product&chain rule?? I've got no idea what you just did BTW!

7. Originally Posted by elpermic
I was working with your way and I still don't get how to do it...

f'(x)= (2x)(√(9-x^2) + (-x/2√(9-x^2))(x^2)
= 2x√(9-x^2) + (-x^3/2√(9-x^2))

Did I do something wrong??
The bit in red should be plus, and the 2 shouldn't be there. This is why:

$\displaystyle f'(x)=\underbrace{(2x)}_{u'(x)}\underbrace{(\sqrt{ 9-x^2})}_{v(x)}+\underbrace{\left(-\frac{x}{\sqrt{9-x^2}}\right)}_{v'(x)}\underbrace{(x^2)}_{u(x)}=2x\ sqrt{9-x^2}+\frac{-x^3}{\sqrt{9-x^2}}$

So we have [as of now] $\displaystyle f'(x)=2x\sqrt{9-x^2}+\frac{-x^3}{\sqrt{9-x^2}}$

To combine the two terms, they need to have a common denominator. In this case, the common denominator is $\displaystyle \sqrt{9-x^2}$.

So we see that $\displaystyle 2x\sqrt{9-x^2}\cdot{\color{red}\frac{\sqrt{9-x^2}}{\sqrt{9-x^2}}}+\frac{-x^3}{\sqrt{9-x^2}}=\frac{2x(9-x^2)-x^3}{\sqrt{9-x^2}}$ $\displaystyle =\frac{18x-2x^3-x^3}{\sqrt{9-x^2}}=\color{red}\boxed{\frac{18x-3x^3}{\sqrt{9-x^2}}}$

8. NOW I understand the problem.. Thanks!