# inverse derivs...

• Oct 5th 2007, 10:30 AM
mathaction
inverse derivs...
http://www.webassign.net/www17/symIm...91c73c1343.gif

http://www.webassign.net/www17/symIm...4906f589c8.gif

Consider the function below where x > 0.

http://www.webassign.net/www17/symIm...1a29f63c63.gif

(a) Find and simplify f '(x).

I need a little help with these bad johnnys...can someone show me how to do them, prefferably the steps...many thanks...

mathaction
• Oct 5th 2007, 01:31 PM
TKHunny
Are you serious about needing help on g(x)? Take a good close look at it. You'll kick yourself.

These are just Chain Rule exercises.

For the first, you need to know these derivatives:

$\displaystyle \frac{d}{dt}\left(\frac{1}{t}\right)$

$\displaystyle \frac{d}{dt}(arctan(t))$

$\displaystyle \frac{d}{dt}(3t)$

Show us what you get.
• Oct 5th 2007, 01:57 PM
mathaction
Quote:

Originally Posted by TKHunny
Are you serious about needing help on g(x)? Take a good close look at it. You'll kick yourself.

These are just Chain Rule exercises.

For the first, you need to know these derivatives:

$\displaystyle \frac{d}{dt}\left(\frac{1}{t}\right)$

$\displaystyle \frac{d}{dt}(arctan(t))$

$\displaystyle \frac{d}{dt}(3t)$

Show us what you get.

yeah i already did the g(x), its just pi because sin and arcsin cancel out and the derivative of pi(x) is just pi cause pi is constant and d/dx is just 1

d/dx(1/t) is 1/t^2, d/dx(arctan(t) is 1/(1+t^2), and d/dx(3t) is 3...now what do i do from here...
• Oct 5th 2007, 02:06 PM
Soroban
Hello, mathaction!

I must assume you know the derivative of the arctangent.

. . $\displaystyle \frac{d}{dx}(\arctan u) \;=\;\frac{u'}{1+u^2}$

Quote:

$\displaystyle f(x)\:=\:\frac{1}{\arctan(3t)}$
We have: .$\displaystyle f(x)\;=\;[\arctan(3t)]^{-1}$

Chain Rule: .$\displaystyle f'(x)\;=\;-1[\arctan(3t)]^{-2}\cdot\frac{1}{1+(3t)^2}\cdot 3 \;=\;\frac{-3}{[\arctan(3t)]^2(1 + 9t^2)}$

Quote:

$\displaystyle g(x)\;=\;\sin\left[\arcsin(\pi x)\right]$
We have: .$\displaystyle g(x)\;=\;\sin\left[\arcsin(\pi x)\right] \;=\;\pi x$

Therefore: .$\displaystyle g'(x) \,=\,pi$

Quote:

$\displaystyle f(x) \;=\;\arctan(x) + \arctan\left(\frac{1}{x}\right)$
This one has a surprising answer . . .

We have: .$\displaystyle f'(x)\;=\;\frac{1}{1+x^2} + \frac{1}{1 + \left(\frac{1}{x}\right)^2}\cdot(-x^{-2}) \;= \;\frac{1}{1+x^2} + \frac{1}{1+\frac{1}{x^2}}\cdot\frac{-1}{x^2}$

. . $\displaystyle = \;\frac{1}{1+x^2} - \frac{1}{x^2+1} \;=\;0$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

How come it works out to zero?

Let $\displaystyle \alpha = \arctan(x)$
Then: .$\displaystyle \tan\alpha \:=\:\frac{x}{1}\:=\:\frac{opp}{adj}$

$\displaystyle \alpha$ is in a right triangle with: $\displaystyle opp = x,\;adj = 1$
Here is the triangle:
Code:

                        *                     *  *                   *    *               *        * x             *          *         * α            *       * * * * * * * * * *                 1

Let $\displaystyle \beta = \arctan\left(\frac{1}{x}\right)$
Then: .$\displaystyle \tan\beta \:=\: \frac{1}{x}\:=\:\frac{opp}{adj}$
$\displaystyle \beta$ is in a right triangle with: $\displaystyle opp = 1,\;adj = x$
Here is the triangle:
Code:

                        *                     * β*                   *    *               *        * x             *          *         *              *       * * * * * * * * * *                 1

Get it?
$\displaystyle \alpha$ and $\displaystyle \beta$ are in the same right triangle.
They are complementary: .$\displaystyle \alpha + \beta \:=\:\frac{\pi}{2}$

We just differentiated a constant function: .$\displaystyle f(x) \:=\:\frac{\pi}{2}$