# Thread: derivatives

1. ## derivatives

I picked up a new textbook and this exercise was in it

 x u(x) v(x) u'(x) v'(x) 2 -3 4 6 5 3 2 7 -2 1

then it wants me to find $\frac{dy}{dx}$ for $\frac{u(x)}{v(x)}$ for x = 2 and $v(u(x))$ when x = 3

the first one was easy and I got $\frac{39}{16}$

the second one I figured would be easy as well. the book has $-10$ but I got

$dy/dx = v'(u(x)) * u'(x) \Rightarrow (1)(2)(-2)=-4$

2. ## Re: derivatives

The book says that $v'(u(x)) = v'(2)$ which is 5 in the table. still unclear about this.

3. ## Re: derivatives

I'm not connecting your table and your workings. Aka. I don't understand what you're supposed to do. Could you explain more clearly?

4. ## Re: derivatives

you take the given values in the table to calculate the derivative asked.

such as,

the first one asked is find dy/dx of u(x)/v(x) when x = 2 which is

$\frac{u'(x)v(x) - v'(x)u(x)}{[v(x)]^2} \Rightarrow \frac{6(4) - 5(-3)}{4^2} \Rightarrow dy/dx = \frac{39}{16}$

for the second one is dy/dx for v(u(x)) when x = 3.

I get $v'(u(x)) * u'(x) = 1(2)*(-2) = -4 = dy/dx$

the book says it is -10 though.

5. ## Re: derivatives

I think I get it $v'(u(x))$ isn't $1(2)$ but instead its $v'(u(x)) \Rightarrow v'(2)$ because $u(x) = 2, when\, x = 3$

6. ## Re: derivatives

I think there's something wrong in your working for x=3 but I can't find what is wrong. Srry!