# Finding third derivative

• Jul 9th 2012, 03:48 PM
daigo
Finding third derivative
Quote:

$y = (5x - 1)^{\frac{1}{2}}$
So to find the 1st derivative I use the Chain Rule:

$y' = \frac{1}{2}(5x - 1)^{\frac{1}{2} - 1} \cdot (5 - 0) \\
= \frac{5}{2}(5x - 1)^{-\frac{1}{2}}$

2nd:

Multiplication rule [f'(x)*g(x) + f(x)*g'(x)] and Chain Rule:

$y'' = \frac{5}{2}(5x - 1)^{-\frac{1}{2}} \\
= 0\cdot (5x - 1)^{-\frac{1}{2}} + \frac{5}{2}\cdot [-\frac{1}{2}(5x - 1)^{-\frac{1}{2} - 1} \cdot (5 - 0)] \\
= \frac{5}{2}\cdot (-\frac{5}{2})(5x - 1)^{-\frac{3}{2}} \\
= -\frac{25}{4}(5x - 1)^{-\frac{3}{2}}$

3rd:

Multiplication rule [f'(x)*g(x) + f(x)*g'(x)] and Chain Rule:

$y''' = -\frac{25}{4}(5x - 1)^{-\frac{3}{2}} \\
= 0\cdot (5x - 1)^{-\frac{3}{2}} + -\frac{25}{4}\cdot [(-\frac{3}{2}(5x - 1)^{-\frac{3}{2}-1}) \cdot (5 - 0)] \\
= -\frac{25}{4}\cdot (-\frac{15}{2})(5x - 1)^{-\frac{5}{2}} \\
= \frac{375}{8}(5x - 1)^{-\frac{5}{2}} \\
= \frac{375}{8(5x-1)^{\frac{5}{2}}}$

I think I got this one wrong but this is how I worked it out...have I done any step incorrectly here?
• Jul 9th 2012, 04:02 PM
skeeter
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

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

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

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

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

$y''' = \frac{75}{8}(5x-1)^{-5/2} \cdot 5 = \frac{375}{8}(5x-1)^{-5/2} = \frac{375}{8(5x-1)^{5/2}}$
• Jul 9th 2012, 04:10 PM
daigo
Re: Finding third derivative
Sorry, I forgot what the rule is called

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

Which one is this
• Jul 9th 2012, 04:13 PM
daigo
Re: Finding third derivative
Quote:

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

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
• Jul 9th 2012, 04:15 PM
pickslides
Re: Finding third derivative
Quote:

Originally Posted by daigo
Sorry, I forgot what the rule is called

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

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)