# Thread: A Few Differentiation Questions

1. ## A Few Differentiation Questions

Hi
I need help on the following questions:

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

This is what i have done:

$\displaystyle x= tan(y)$

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

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

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

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

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

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

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

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

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

What have i done wrong??

P.S

2. For the first 1.

$\displaystyle x= tan(y)$

Differentiating wrt $\displaystyle x$:

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

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

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

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

3. Originally Posted by craig
For the first 1.

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

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

Since $\displaystyle x= tan(y)$, $\displaystyle \frac{dy}{dx} = \frac{1}{x^2 + 1}$
How did you get $\displaystyle tan^2y + 1$??

4. don't worry i know now, its from the identity $\displaystyle sec^2(x) = 1+ tan^2(x)$

5. can anyone answer my other question?

6. 3) Find the derivative of [tex]e^{-2x}sinh{5x}

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

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

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

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

7. yep thats correct, thanks

8. can anyone help me on question 2)

I have tried it again and i still get the same answer

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

10. 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
$\displaystyle \frac{2}{\sqrt{9-4x^2}}$, 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?

11. 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 $\displaystyle \frac{d}{dx}\left( \frac{1}{x\sqrt{\frac{1}{x^2}}}\right)$. YOu can of course rewrite it in it's current form to be $\displaystyle x^{-1}\left(\frac{1}{x^2}\right)^{-\frac{1}{2}}$ and use something like the product rule or the chain rule to solve, which would be long and complicated. Or you could note that $\displaystyle \frac{1}{x\sqrt{\frac{1}{x^2}}}$ 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.