# Finding third derivative

#### daigo

##### Member
$$y = (5x - 1)^{\frac{1}{2}}$$
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?

Staff member

#### daigo

##### Member
Oh wow, I typed in "third derivative of" and then the given equation and it found the exact answer I got lol

If only my HP-15c could do this too

Do you know if graphing calculators are capable of this?

#### Jameson

Staff member
Certain ones are, for example the TI-89 but it's better to do these calculations without this kind of help. With great power comes great responsibility!

Many calculators can calculate the derivative at a certain point, so you could type in something like Deriv(x^2, x, 2) meaning calculate the derivative of x^2 with respect to x at the point x=2 and it would output 4. Far less calculators can output the general derivative.

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#### CaptainBlack

##### Well-known member
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?
You can elliminate a lot of the algebra when doing this by observing that:
$y'=\frac{5}{2}\;y^{-1}$
so:
$y''=-\;\frac{5}{2}y'\;y^{-2}=-\frac{25}{4}\;y^{-3}$
Then differentiating again gives:
$y'''=\frac{25\times 3 \times 5}{8}y^{-5}$

CB

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