Re: Finding third derivative

don't make it so hard on yourself ... on every iteration of a derivative, the derivative of (5x-1) becomes an additional factor

$\displaystyle y = (5x-1)^{1/2}$

$\displaystyle y' = \frac{1}{2}(5x-1)^{-1/2} \cdot 5 = \frac{5}{2}(5x-1)^{-1/2}$

you don't need the product rule ... $\displaystyle \frac{d}{dx} \left[k \cdot f(x)\right] = k \cdot f'(x)$

$\displaystyle y'' = -\frac{5}{4}(5x-1)^{-3/2} \cdot 5 = -\frac{25}{4}(5x-1)^{-3/2}$

$\displaystyle y''' = \frac{75}{8}(5x-1)^{-5/2} \cdot 5 = \frac{375}{8}(5x-1)^{-5/2} = \frac{375}{8(5x-1)^{5/2}}$

Re: Finding third derivative

Sorry, I forgot what the rule is called

$\displaystyle \frac{d}{dx}[f(x) \cdot g(x)] = [\frac{d}{dx}f(x) \cdot g(x)] + [f(x) \cdot \frac{d}{dx}g(x)]$

Which one is this

Re: Finding third derivative

Quote:

Originally Posted by

**skeeter** you don't need the product rule ... $\displaystyle \frac{d}{dx} \left[k \cdot f(x)\right] = k \cdot f'(x)$

Oh wow, the use of this basic rule was very clever. If I had thought of it it would've made this so much easier

Re: Finding third derivative

Quote:

Originally Posted by

**daigo** Sorry, I forgot what the rule is called

$\displaystyle \frac{d}{dx}[f(x) \cdot g(x)] = [\frac{d}{dx}f(x) \cdot g(x)] + [f(x) \cdot \frac{d}{dx}g(x)]$

Which one is this

It is the product rule b/c you are taking the derivative of a product i.e f(x) times g(x)