Derivative

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February 10th 2010, 08:41 PM
Noxide

Derivative

y=|x|(tanx)^(1/2)

find y'
February 10th 2010, 09:03 PM
General

Quote:

Originally Posted by Noxide

y=|x|(tanx)^(1/2)

find y'

Are you given a specific values for x?
or an interval?
February 10th 2010, 09:24 PM
lgstarn

Quote:

Originally Posted by Noxide

y=|x|(tanx)^(1/2)

find y'

Run the online derivative calculator:

webMathematica Explorations: Step-by-Step Derivatives

Derivative of: x*Tan[x]^(1/2)
With respect to: x

Gives:

$\sqrt{\tan(x)} + \frac{x \sec^2(x)}{2\sqrt{\tan(x)}}$

This is valid for $x > 0$

For $x < 0$ , you have

Derivative of: -x*Tan[x]^(1/2)
With respect to: x

which gives:

$\sqrt{\tan(x)} - \frac{x \sec^2(x)}{2\sqrt{\tan(x)}}$

So we can combine these two into one expression for $x \ne 0$ :

$\sqrt{\tan(x)} + \left|\frac{x \sec^2(x)}{2\sqrt{\tan(x)}}\right|$
February 11th 2010, 09:15 AM
Noxide

thanks, i was on the right track
February 11th 2010, 10:17 AM
Noxide

is it correct to say (with respect to my first question):

if x = 0 then y = 0 and d0/dx = 0?
or should i differentiate first using the positive case of x then input the value of x into the derivative and get 0?
February 12th 2010, 07:49 AM
lgstarn

Quote:

Originally Posted by Noxide

is it correct to say (with respect to my first question):

if x = 0 then y = 0 and d0/dx = 0?
or should i differentiate first using the positive case of x then input the value of x into the derivative and get 0?

Well, if $y$ is a function of $x$ , in other words, $y = f(x)$ , you can't do that directly. In this case,

$y(x)=|x|\sqrt{\tan x}$

So we compute the derivative as before to find $y'(x) = \frac{dy}{dx}(x)$ . Only once you've done that can you put in $x = 0$ to find what the value of $y'(0)$ is.

But because you have an absolute value, at $x = 0$ the derivative doesn't exist. That's because the slope of the tangent line of $|x|$ changes sign at $x=0$ .
February 13th 2010, 12:11 PM
Noxide

Thanks.

I can picture that there would be an infinite number of tangent lines at that point. So there are infinite derivatives. Is there a way to get a general equation for those infinite derivatives?
February 13th 2010, 12:44 PM
drumist

Quote:

Originally Posted by Noxide

Thanks.

I can picture that there would be an infinite number of tangent lines at that point. So there are infinite derivatives. Is there a way to get a general equation for those infinite derivatives?

We just say it isn't differentiable at x=0 and exclude the point x=0 from the domain of the derivative. You don't need to bother doing anything else.

The domains of the function and its derivative are a bit complicated though (since $\sqrt{\tan x}$ only exists where $\tan x \ge 0$ ). Are you asked to state the domain of the derivative?
February 13th 2010, 01:06 PM
lgstarn

Quote:

Originally Posted by Noxide

Thanks.

I can picture that there would be an infinite number of tangent lines at that point. So there are infinite derivatives. Is there a way to get a general equation for those infinite derivatives?

Actually, you are correct! That is the concept of a "weak derivative." This is a bit advanced though. You wouldn't have to see that stuff until you became a graduate student or a senior working in mathematics.

Don't let the following links blow your mind - just be aware that what you are talking about does exist.

Weak derivative - Wikipedia, the free encyclopedia
Subderivative - Wikipedia, the free encyclopedia