find second derivative?

Tom2013

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First derivative is y=f(x) is y'=(x-2)^-2/3. Find the second derivative (y"). I think the answer is -2/3(x-2)^-5/3, is this correct? If not how do I get the power outside the parentheses?
Thanks
 
First derivative is y=f(x) is y'=(x-2)^-2/3. Find the second derivative (y"). I think the answer is -2/3(x-2)^-5/3, is this correct? If not how do I get the power outside the parentheses?
Thanks
\(\displaystyle \displaystyle y\prime = (x-2)^{-2/3}\)

\(\displaystyle \displaystyle y\prime \prime = -\left( \dfrac{2}{3} \right) (x-2)^{-5/3} \)

is correct. The power of the expression in parentheses, (x - 2)^n, is differentiated by the chain rule, but that doesn't show because the derivative of (x-2) is 1.


EDIT: this post does not seem to qualify as "News." Perhaps it should be moved to the "Calculus" folder?
 
Last edited:
Yes, that is correct. The derivative of \(\displaystyle x^n\) is \(\displaystyle nx^{n-1}\). The second derivative is just the derivative of the first derivative. Here, what y itself is is not important. with n= -2/3, n- 1= -2/3- 3/3= -5/3 so the second derivative is \(\displaystyle (-2/3)(x- 2)^{-5/3}\)
 
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