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Finding third derivative

daigo

Member
Jun 27, 2012
60
[tex]y = (5x - 1)^{\frac{1}{2}}[/tex]
So to find the 1st derivative I use the Chain Rule:

[tex]y' = \frac{1}{2}(5x - 1)^{\frac{1}{2} - 1} \cdot (5 - 0) \\
= \frac{5}{2}(5x - 1)^{-\frac{1}{2}}[/tex]

2nd:

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

[tex]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}}[/tex]

3rd:

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

[tex]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}}}[/tex]

I think I got this one wrong but this is how I worked it out...have I done any step incorrectly here?
 

daigo

Member
Jun 27, 2012
60
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

Administrator
Staff member
Jan 26, 2012
4,041
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
Jan 26, 2012
890
So to find the 1st derivative I use the Chain Rule:

[tex]y' = \frac{1}{2}(5x - 1)^{\frac{1}{2} - 1} \cdot (5 - 0) \\
= \frac{5}{2}(5x - 1)^{-\frac{1}{2}}[/tex]

2nd:

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

[tex]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}}[/tex]

3rd:

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

[tex]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}}}[/tex]

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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