anonymous
  • anonymous
y=e^sqrt(x)) x=1 Evaluate dy/dx, meaning they want me to use chain rule, but I cant get the right answer, help!!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
johnweldon1993
  • johnweldon1993
\[\large \frac{dy}{dx} = \frac{d(e^{\sqrt{x}})}{dx}\] The chain rule says...take the derivative of the inside...and multiply it to the derivative of the outside... Take the "inside" to be \(\large \sqrt{x} \)
johnweldon1993
  • johnweldon1993
So if we treat \(\large u = \sqrt{x}\) ... then \(\large y = e^{u}\) So the derivative of \(\large \sqrt{x}\) is \(\large \frac{1}{2\sqrt{x}}\) and the derivative of \(\large e^{u} = e^{u}\) So when you multiply them together you get \[\large \frac{e^{u}}{2\sqrt{x}}\] But of course you want to go back and replace what 'u' was...so our final answer would be \[\large \frac{e^{\sqrt{x}}}{2\sqrt{x}}\]
anonymous
  • anonymous
So simple but somehow I managed to mess this on up, maybe the fact that Ive been studying since 6 am and now its almost 4 pm. Thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

johnweldon1993
  • johnweldon1993
Lol yeah after 2 hours I'm all set...should probably take a break!

Looking for something else?

Not the answer you are looking for? Search for more explanations.