physics604
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1. Find the derivative of y=e[itex]\sqrt{x}[/itex]
Chain rule
y=eu
[itex]\frac{dy}{du}[/itex]= ueu-1
u=[itex]\sqrt{x}[/itex]
[itex]\frac{du}{dx}[/itex]= [itex]\frac{1}{2}[/itex]x-1/2
[itex]\frac{dy}{dx}[/itex]= [itex]\sqrt{x}[/itex]e[itex]\sqrt{x}[/itex]-1 × [itex]\frac{1}{2}[/itex]x-1/2
= [itex]\sqrt{x}[/itex] [itex]\frac{e^\sqrt{x}}{e}[/itex] × [itex]\frac{1}{2\sqrt{x}}[/itex]
= [itex]\frac{e^\sqrt{x}}{2e}[/itex]
The answer to this question is [itex]\frac{e^\sqrt{x}}{2\sqrt{x}}[/itex]. What did I do wrong?
Any help is much appreciated.
Homework Equations
Chain rule
The Attempt at a Solution
y=eu
[itex]\frac{dy}{du}[/itex]= ueu-1
u=[itex]\sqrt{x}[/itex]
[itex]\frac{du}{dx}[/itex]= [itex]\frac{1}{2}[/itex]x-1/2
[itex]\frac{dy}{dx}[/itex]= [itex]\sqrt{x}[/itex]e[itex]\sqrt{x}[/itex]-1 × [itex]\frac{1}{2}[/itex]x-1/2
= [itex]\sqrt{x}[/itex] [itex]\frac{e^\sqrt{x}}{e}[/itex] × [itex]\frac{1}{2\sqrt{x}}[/itex]
= [itex]\frac{e^\sqrt{x}}{2e}[/itex]
The answer to this question is [itex]\frac{e^\sqrt{x}}{2\sqrt{x}}[/itex]. What did I do wrong?
Any help is much appreciated.
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