The second derivative

  • Thread starter FChebli
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  • #1
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How do I find the second derivative of the function:

f(x)= x^(2/3) (6-x)^(1/3)

I have found the first derivative and checked my solution:

′()= 4− / ^(1/3) (6−)^(2/3)


The final solution is supposed to be:

''()= -8 / ^(4/3) (6−)^(5/3)

I know almost all the steps but I couldn't reach the final answer! Can you please help me?
Thanks in advance!
 

Answers and Replies

  • #2
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but I couldn't reach the final answer!
What did you get?
Which differentiation rules do you know?
It is impossible to see what you did wrong if you don't show your work.

I think your f'(x) and f''(x) are missing brackets.
 
  • #3
lurflurf
Homework Helper
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let
$$\mathrm{f}(x)=\mathrm{u}(x)\mathrm{v}(x) \\
\text{with} \\
\mathrm{u}(x)=x^{2/3} \\
\mathrm{v}(x)=(6-x)^{1/3} $$
By the product rule
$$\mathrm{f}^{\prime \prime}(x)=\mathrm{u}^{\prime \prime}(x)\mathrm{v}(x)+2\mathrm{u}^{ \prime}(x)\mathrm{v}^{\prime }(x)+\mathrm{u}(x)\mathrm{v}^{\prime \prime}(x)$$
what did you find for
u'(x)
u''(x)
v'(x)
v''(x)
?
 
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  • #4
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let
$$\mathrm{f}(x)=\mathrm{u}(x)\mathrm{v}(x) \\
\text{with} \\
\mathrm{u}(x)=x^{2/3} \\
\mathrm{v}(x)=(6-x)^{1/3} $$
By the product rule
$$\mathrm{f}^{\prime \prime}(x)=\mathrm{u}^{\prime \prime}(x)\mathrm{v}(x)+2\mathrm{u}^{ \prime}(x)\mathrm{v}^{\prime }(x)+\mathrm{u}(x)\mathrm{v}^{\prime \prime}(x)$$
?

Where did you get this formula from? How did you wind up to this product rule?

I got u'(x) = (2/3) x^(-1/3)
u''(x) = (-2/9) x^(-4/3)

v'(x) = -(1/3) < (6-x)^(-2/3) >
v''(x) = -(2/9) < (6-x)^(-5/3) >


using your formula it's difficult to simplify after this step

< (-2/9) (x^(-4/3)) ((6-x)^(1/3)) > + < (-4/9) (x^(-1/3)) ((6-x)^(-2/3)) > + < (-2/9) (x^(2/3)) ((6-x)^(-5/3)) >
 
  • #5
35,801
12,522
lurfturf used the product rule both for f(x) and then for the f'(x) you get there.

What are those < >?
v'(x) has a wong prefactor.

To simplify, you need exponentiation rules, especially a^(b+c)=...
 
  • #6
lurflurf
Homework Helper
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good, to simplify factor


$$\mathrm{f}^{\prime \prime}(x)=\mathrm{u}^{\prime \prime}(x)\mathrm{v}(x)+2\mathrm{u}^{ \prime}(x)\mathrm{v}^{\prime }(x)+\mathrm{u}(x)\mathrm{v}^{\prime \prime}(x)
\\
=\left( x^{-2/3} \right)^{\prime \prime} (6-x)^{1/3}+2 \left( x^{-2/3}\right)^{\prime } \left( (6-x)^{-1/3}\right)^{\prime }+ x^{-2/3}\left( (6-x)^{-1/3}\right)^{\prime \prime}
\\
=\left( -\frac{2}{9} x^{-4/3}\right) \left( (6-x)^{1/3}\right)+2\left( \frac{2}{3} x^{-1/3} \right) \left( -\frac{1}{3}(6-x)^{-2/3}\right)+\left( x^{2/3}\right) \left( -\frac{2}{9}(6-x)^{-5/3} \right)
\\
-\frac{2}{9} x^{-4/3} (6-x)^{-5/3}((6-x)^2+2x(6-x)+x^2)
\\
=-\frac{2}{9}x^{-4/3}(6-x)^{-5/3}((6-x)+x)^2
$$
and so on
 

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