I can't understand where the 2 comes from in the following transformation. $$\frac{\partial}{\partial k}f(k)=\frac{\partial}{\partial k}k^{0.5}=\frac{1}{2k^{0.5}}$$ Any help would be appreciated.
Asked
Active
Viewed 61 times
3
-
HINT: if $y=x^n$ then $dy/dx=nx^{n-1}$. Look at what this gives for $n=0.5$. – Mufasa Aug 21 '14 at 11:52
2 Answers
3
Having looked at your previous question, I see the issue here. The key point is that we are computing a partial derivative here. $\partial$ is not a number that gets canceled out.
The derivative of $f(k) = k^{\frac12}$ is \begin{equation*} f'(k) = \frac12 k^{-\frac12}. \end{equation*}.
littleO
- 51,938
-
Wow. I would've never pinpointed that issue with OP so easily! +1 for the great answer. – Patrick Da Silva Aug 21 '14 at 12:04
0
First note that $$\frac{\partial}{\partial k}k^n=nk^{n-1}$$ So here's how the $2$ appears $$ \frac{\partial}{\partial k}k^{0.5}= \frac{\partial}{\partial k}k^{\frac{1}{2}}=\frac{1}{2}k^{\frac{1}{2}-1} $$ $$ = \frac{1}{2}k^{-\frac{1}{2}}= \frac{1}{2k^{\frac{1}{2}}}=\frac{1}{2\sqrt{k}} $$
k170
- 9,045