e to the x and square root problem
- Thread starter Jason76
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What is your question? And why did the denominator change between the second and third lines above?\(\displaystyle f'(x) = 2e^{x} + \dfrac{8}{\sqrt[3]{x}}\)
\(\displaystyle f'(x) = 2e^{x} + \dfrac{8}{x^{1/3}}\)
\(\displaystyle f'(x) = 2e^{x} + \dfrac{8}{1/3x^{-2/3}}\)
Please be complete. Thank you!
D
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f'(x) = \(\displaystyle 2e^{x} + \dfrac{8}{\sqrt[3]{x}}\)
f'(x) = \(\displaystyle 2e^{x} + \dfrac{8}{x^{1/3}}\)
f'(x) = \(\displaystyle 2e^{x} + \dfrac{8}{1/3x^{-2/3}}\)
Is that correct?
Last edited:
D
Deleted member 4993
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Post Edited
\(\displaystyle f(x) = 2e^{x} + \dfrac{8}{\sqrt[3]{x}}\)
f'(x) = \(\displaystyle 2e^{x} + \dfrac{8}{x^{1/3}}\)................. why is that f'(x)?
\(\displaystyle f'(x) = 2e^{x} + \dfrac{8}{1/3x^{-2/3}}\)
Cannot do that!!!
\(\displaystyle f(x) = 2e^{x} + \dfrac{8}{\sqrt[3]{x}}\)
\(\displaystyle f(x) = 2e^{x} + 8*{x}^{\frac{-1}{3}}\)
\(\displaystyle f'(x) = 2e^{x} + 8*\left [\dfrac{-1}{3}{x}^{(\frac{-1}{3}-1)}\right ]\)
\(\displaystyle f'(x) = 2e^{x} - \dfrac{8}{3}{x}^{-\frac{4}{3}}\)
These problems are very simple if you would just pay attention.....
HallsofIvy
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You have repeatedly written f'(x) when you have NOT YET differentiated! You need to pay more attention to what you are writing.
\(\displaystyle f'(x) = 2e^{x} + \dfrac{8}{1/3x^{-2/3}}\)
Last edited:
D
Deleted member 4993
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.Notation problem fixed, and also original post edited.
\(\displaystyle f(x) = 2e^{x} + \dfrac{8}{\sqrt[3]{x}}\)
\(\displaystyle f(x) = 2e^{x} + \dfrac{8}{x^{1/3}}\)
\(\displaystyle f(x) = 2e^{x} + \dfrac{8}{1/3x^{-2/3}}\)