suxatphysix said:
soooo 1/2 \frac{z-1}{z+1}^{-1/2} (1) ?
No, that's
not correct.
Why is it (1)? As others have pointed out, you should use
The Chain Rule:
\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}, i.e the derivative of f with respect to x is the derivative of f with respect to u times the derivative of u with respect to x. (u is any function that you choose, so that the problem becomes easier). There are problems that you have to apply the rules twice, threefold, or even more:
\frac{df}{dx} = \frac{df}{du_1} \times \frac{du_1}{dx} = \frac{df}{du_1} \times \left( \frac{du_1}{du_2} \times \frac{du_2}{dx} \right) = \frac{df}{du_1} \times \frac{du_1}{du_2} \times \left( \frac{du_2}{du_3} \times \frac{du_3}{dx} \right) = ...
Example:
Find the derivative of \tan \left( \sqrt{\cos x} \right) with respect to x.
-------------------
Now that looks monstrous. But if you let u = \sqrt{\cos x}, the whole thing becomes: tan(u).
Apply
The Chain Rule here, we have:
\tan \left( \sqrt{\cos x} \right)'_x = \tan \left( u \right)'_x = \tan \left( u \right)'_u \times u'_x = \frac{1}{\cos ^ 2 u} \times (\sqrt{\cos x})'_x
Apply the Rule once more, let t = cos(x)
... = \frac{1}{\cos ^ 2 u} \times (\sqrt{\cos x})'_x = \frac{1}{\cos ^ 2 u} \times (\sqrt{t})'_x = \frac{1}{\cos ^ 2 u} \times (\sqrt{t})'_t \times t'_x = \frac{1}{\cos ^ 2 u} \frac{1}{2 \sqrt{t}} (\cos x)'_x = -\frac{1}{\cos ^ 2 u} \frac{1}{2 \sqrt{t}} \sin x
Now, change u, and t back to x's, we have:
... = -\frac{1}{\cos ^ 2 u} \frac{1}{2 \sqrt{t}} \sin x = -\frac{1}{\cos ^ 2 (\sqrt{\cos x})} \times \frac{1}{2 \sqrt{\cos x}} \sin x.
To make it look a little bit
nicer. You can pull the 1/2 factor in front of the whole expression, like this:
... = - \frac{1}{2} \times\frac{1}{\cos ^ 2 (\sqrt{\cos x})} \times \frac{\sin x}{\sqrt{\cos x}}It's done. :)
Can you do the same to your 2 problems? :)