Question

help? ;D

Answer 1

\[\frac{ e^\sqrt{x} }{ 2\sqrt{x} }-\sec^2(x)\]
Finding dy/dx

Answer 2

You've a Quotient Rule and some Chain Rules in your future. Give it a go and we'll see if you wander off.

Answer 3

Yeah, I'm trying it out right now

Answer 4

I agree quotient rule on the left and use the chain rule on the right\[f(x)=\sec x\]\[f'(x)=\sec x \tan x\]Visualize the right as \[(\sec x)^2\]

Answer 5

\[f(x)=e^x\]\[f'(x)=e^xx'\]

Answer 6

It's NOT pretty. \(\dfrac{2\cdot\sqrt{x}\cdot \dfrac{d}{dx}e^{\sqrt{x}} - e^{\sqrt{x}}\cdot\dfrac{d}{dx}2\cdot\sqrt{x}}{\left(2\cdot\sqrt{x}\right)^{2}}\) and that just the left piece!

Sorry, I was just having fun coding that.

Answer 7

Yeah, I finished the quotient rule part. I'm working on the chain rule

Answer 8

oh my lordy!!! I'm doing the wrong question!

it's \[e^{\sqrt{x}}-\tan(x)\]

Answer 9

hah :D

Answer 10

That's hilarious, the thing you posted at the start is actually the answer lolol.
So you posted the answer instead of the question? XD

Answer 11

I was finding the derivative of the flippen answer to the question... erghh But, i got it.

Answer 12

I need sleep haha!

Answer 13

too funny XD