Differentiate e^x and Trig Functions

gabyoh23
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Homework Statement


Differentiate
e^x * cotx / 5sqrtx^2
[Sorry for not using the formatting things. They didn't seem to be working for me, and this is urgent!]


Homework Equations


The quotient rule seems like that's the way to go...


The Attempt at a Solution


At first I tried using the product rule on the numerator, then plugging that into the quotient rule formula, but that was needlessly complicated. So, I went straight into using the quotient rule, but I got a huge messy equation. Could anyone clarify what I SHOULD be getting?

All help is greatly appreciated!
 
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Both ways are correct. Either do product rule inside of Quotient or do Product then do Quotient. Both will be potentially messy.
 
also is that 5sqrt(x^2) or (5sqrt(x))^2 or what?
because that should simple things out for you.
 
It's 5sqrt(x^2).
Sorry about that.
 
The quotient rule is never worth remembering IMO. Just use the product rule and think of the derivative of a quotient as

<br /> \frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{d}{dx}\left(f(x) \ g(x)^{-1}\right)<br />

and don't forget to apply the chain rule when differentiating g(x)^{-1}.

It's too easy to forget the quotient rule on an exam, and also too easy to screw it up when you're in a rush to get everything done in 50 minutes on a midterm. The product rule and chain rule are easy though, and critical to know anyways.
 
What's the square root of x^2?

that will make it a little simpler.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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