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differentiate sqrtx
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i got (1/2)sqrtx
The power rule \[\Large \frac{d}{dx}x^n=nx^{n-1}\] applies even in the most jarring of exponents (for as long as they are constant)
so i am right
Keep that, and the fact that \[\Large \sqrt x = x^{\frac12}\] in mind and the answer will be clear ^_^
1/2sqrtx
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\[\sqrt{x}/2\]
or would it be -sqrtx/2
That's not it :) From the power rule, you should get \[\Large \frac{d}{dx}\sqrt x = \frac{d}{dx}x^{\frac12}=\frac12x^{-\frac12} \] And just simplify
so it would be (1/2)(1/sqrtx)
Yes, or simply put, \[\Large \frac1{2\sqrt x}\] as @vjordillo has put it.
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