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

find y'

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- Feb 10th 2010, 08:41 PMNoxideDerivative
y=|x|(tanx)^(1/2)

find y' - Feb 10th 2010, 09:03 PMGeneral
- Feb 10th 2010, 09:24 PMlgstarn
Run the online derivative calculator:

webMathematica Explorations: Step-by-Step Derivatives

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

With respect to: x

Gives:

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

This is valid for $\displaystyle x > 0$

For $\displaystyle x < 0$, you have

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

With respect to: x

which gives:

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

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

$\displaystyle \sqrt{\tan(x)} + \left|\frac{x \sec^2(x)}{2\sqrt{\tan(x)}}\right|$ - Feb 11th 2010, 09:15 AMNoxide
thanks, i was on the right track

- Feb 11th 2010, 10:17 AMNoxide
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? - Feb 12th 2010, 07:49 AMlgstarn
Well, if $\displaystyle y$ is a function of $\displaystyle x$, in other words, $\displaystyle y = f(x)$, you can't do that directly. In this case,

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

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

But because you have an absolute value, at $\displaystyle x = 0$ the derivative doesn't exist. That's because the slope of the tangent line of $\displaystyle |x|$ changes sign at $\displaystyle x=0$. - Feb 13th 2010, 12:11 PMNoxide
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? - Feb 13th 2010, 12:44 PMdrumist
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 $\displaystyle \sqrt{\tan x}$ only exists where $\displaystyle \tan x \ge 0$). Are you asked to state the domain of the derivative? - Feb 13th 2010, 01:06 PMlgstarn
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