1. ## Derivative algebra

Problem below:

Explain to me how $\frac{1}{2}\sqrt{-x}(-1)-1$ becomes $\frac{-1}{2\sqrt{-x}}-1$

2. ## Re: Derivative algebra

Originally Posted by Greymalkin
Problem below:
The above work is all correct.
Clearly there is no solution: a positive cannot equal a negative.

3. ## Re: Derivative algebra

I understand that, I just want to know how they made the jump from $\frac{1}{2}\sqrt{-x}(-1)-1$ to $\frac{-1}{2\sqrt{-x}}-1$

4. ## Re: Derivative algebra

Originally Posted by Greymalkin
I understand that, I just want to know how they made the jump from $\frac{1}{2}\sqrt{-x}(-1)-1$ to $\frac{-1}{2\sqrt{-x}}-1$
The derivative of $\sqrt{-x}-x$ is $\frac{-1}{2\sqrt{-x}}-1$. They didn't make a jump.

5. ## Re: Derivative algebra

I see that, but I do not understand how that is done algebraically,
I get $\frac{1}{2}\sqrt{-x}(-1)-1$

$\frac{-1\sqrt{-x}}{2}-1$

$-\frac{\sqrt{-x}}{2}-1$
Which is as far as I get when trying to figure out how it becomes $\frac{-1}{2\sqrt{-x}}-1$

Maybe $\frac{x}{2\sqrt{-x}}-1$ ? closer, but where does the factor of x go?

6. ## Re: Derivative algebra

Think of it like this.

$y=\sqrt{f(x)}$

$y^{\prime}=\frac{f^{\prime}(x)}{2\sqrt{f(x)}}$

Then for your other parts, differentiate like normal.

7. ## Re: Derivative algebra

Originally Posted by Greymalkin
I see that, but I do not understand how that is done algebraically,
I get $\frac{1}{2}\sqrt{-x}(-1)-1$

$\frac{-1\sqrt{-x}}{2}-1$

$-\frac{\sqrt{-x}}{2}-1$
Which is as far as I get when trying to figure out how it becomes $\frac{-1}{2\sqrt{-x}}-1$

Maybe $\frac{x}{2\sqrt{-x}}-1$ ? closer, but where does the factor of x go?
$-\frac{\sqrt{-x}}{2}-1\not=\frac{-1}{2\sqrt{-x}}-1$
It is the case that the are not equal.
What makes you think they are?

8. ## Re: Derivative algebra

Think I just forgot to use brackets over a root symbol when differentiating, because my work did not include ^(-1/2) which easily works with the above. I always seem to fudge the easy stuff in math.

9. ## Re: Derivative algebra

The problem is NOT the differentiation. I suspect that you did not tell us the whole problem! After differentiating they set the derivative equal to 0 and then started solving for x. That would be a natural thing to do if the problem were not just to find the derivative but to find maximum and minimum values for the function.