# A Few Differentiation Questions

• Apr 22nd 2010, 04:04 AM
Paymemoney
A Few Differentiation Questions
Hi
I need help on the following questions:

1)Prove that: $\frac{d}{dx}(arctan(x)) = \frac{1}{1+x^2}$

This is what i have done:

$x= tan(y)$

$\frac{dy}{dx} = \frac{1}{\frac{dy}{dx}} = \frac{1}{sec^2(y)}
$

$sec^2(y) = \frac{1}{cos^2(y)}$

$= \sqrt{1 - sin^2y}$ stuck at this part

2) Find the derivative of $arcsin(\frac{2x}{3})$

This is what i have done:
$\frac{\frac{2}{3}}{\sqrt{1-\frac{4x^2}{9}}}$

$\frac{2}{3\sqrt{1-\frac{4x^2}{9}}}$

3) Find the derivative of $e^{-2x}sinh(5x)$

$\frac{dy}{dx} = \frac{e^{x}xcosh(x) + sinh(x) - xsinhx * e^{x}}{(e^x)^2}$

$\frac{dy}{dx} = \frac{e^{x}(xcosh(x) + sinh(x) - xsinhx)}{(e^x)^2}$

What have i done wrong??

P.S
• Apr 22nd 2010, 04:40 AM
craig
For the first 1.

$x= tan(y)$

Differentiating wrt $x$:

$sec^2y\frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{sec^2y}$

$\frac{dy}{dx} = \frac{1}{tan^2y + 1}$

Since $x= tan(y)$, $\frac{dy}{dx} = \frac{1}{x^2 + 1}$
• Apr 22nd 2010, 02:06 PM
Paymemoney
Quote:

Originally Posted by craig
For the first 1.

$\frac{dy}{dx} = \frac{1}{sec^2y}$

$\frac{dy}{dx} = \frac{1}{tan^2y + 1}$

Since $x= tan(y)$, $\frac{dy}{dx} = \frac{1}{x^2 + 1}$

How did you get $tan^2y + 1$??
• Apr 22nd 2010, 02:15 PM
Paymemoney
don't worry i know now, its from the identity $sec^2(x) = 1+ tan^2(x)$
• Apr 22nd 2010, 04:53 PM
Paymemoney
can anyone answer my other question?
• Apr 22nd 2010, 05:38 PM
Diemo
3) Find the derivative of [tex]e^{-2x}sinh{5x}

Use the cain rule. Lut u = $e^{-2x}$, and v = sinh(5x). Then

$\frac{dy}{dx} =\frac{du}{dx} v+u \frac{dv}{dx}$

$=-2e^{-2x} sinh(5x)+5cosh(5x)e^{-2x}$

$=\frac{5cosh(5x)-2sinh(5x)}{e^{2x}}$

• Apr 22nd 2010, 11:19 PM
Paymemoney
yep thats correct, thanks
• Apr 22nd 2010, 11:24 PM
Paymemoney
can anyone help me on question 2)

I have tried it again and i still get the same answer
• Apr 23rd 2010, 05:25 AM
Diemo
The reason you get the same answer on question two is because it is the right answer. According to wolframalpha anyhow.

Possibly it is in the form
$\frac{2}{\sqrt{9-4x^2}}$, where all I have done there is takke the 3 inside of the square root and simplify.
• Apr 23rd 2010, 02:18 PM
Paymemoney
Quote:

Originally Posted by Diemo
The reason you get the same answer on question two is because it is the right answer. According to wolframalpha anyhow.

Possibly it is in the form
$\frac{2}{\sqrt{9-4x^2}}$, where all I have done there is takke the 3 inside of the square root and simplify.

but why would you want to do that?
• Apr 24th 2010, 07:08 AM
Diemo
Why would you simplify? There are a few reasons. Mainly if you are using it in some future equation though. It makes it much much easier to write, so if you are actually using the result in a larger equation it makes it easier to keep track of things.
Secondly, while it doesn't matter in your case, consider the derivative of $\frac{d}{dx}\left( \frac{1}{x\sqrt{\frac{1}{x^2}}}\right)$. YOu can of course rewrite it in it's current form to be $x^{-1}\left(\frac{1}{x^2}\right)^{-\frac{1}{2}}$ and use something like the product rule or the chain rule to solve, which would be long and complicated. Or you could note that $\frac{1}{x\sqrt{\frac{1}{x^2}}}$ is just 1, hence the derivative is zero. and so on and so forth.

It is a good habit to get into to simplify as much as possible before doing the work. As Skeeter says, work smart not hard, which is something that applied mathematicians (like me) believe in. I once had a lecturer who found out that the place that we had our class was the other side of campus (maybe a ten minute walk) and was too lazy to walk out that far (possibly had other classes to be fair), so he moved the room, and we spent the entire semester squashed up in a room that was waaay to small to hold the entire class. Be lazy about things.