# Find the derivative of (t^2 - 4/t^4)*t^3

1. Oct 16, 2007

### carbz

I have two of these here...

1. The problem statement, all variables and given/known data
Find the derivative

2. Relevant equations
$$(t^2-\frac{4}{t^4})*t^3$$

3. The attempt at a solution
This is how far I got:
$$(t^2-4t(^-4))(t(^3))$$
$$(t^2-4t(^-4))(3t(^2))+(t^3)(2t+16t(^-5))$$

1. The problem statement, all variables and given/known data
Find the derivative

2. Relevant equations
$$f(x) = \frac{(6x+5)(x^3-2)}{(3x^2-5)}$$

3. The attempt at a solution
This is what I got only:
$$(6x+5)(x^3-2)(3x^2-5)(^-1)$$

2. Oct 16, 2007

### rock.freak667

Well for $$(t^2-\frac{4}{t^4})*t^3$$

you could just multiply it out and get $$t^5+\frac{4}{t}=t^5+4t^{-1}$$ and proceed to differentiate w.r.t. t

for the 2nd one
$$\frac{d}{dx}(uv)=\frac{v\frac{du}{dx}+u\frac{dv}{dx}}{v^2}$$

by that formula you should see that the end derivative would be a fraction

3. Oct 16, 2007

### rocomath

always simplify from the beginning if you can

for 2. take the ln of both sides and expand it then take the derivative

4. Oct 18, 2007

### carbz

Allright, thanks. I got both problems worked out correctly now.