1. ## differentiating arcsinx

hi, just have a few questions

use chain rule to find d/dx (1/arcsin(x^2))
i know d/dx (arcsinx)= 1/(1-x^2)^1/2

also for x in domain [-1,1] prove that arcsinx + arccosx = pi/2
for this i took the derivatives of the LHS whiched equaled zero but dont know where to go from here

thanks

2. Originally Posted by b0mb3rz
hi, just have a few questions

use chain rule to find d/dx (1/arcsin(x^2))
i know d/dx (arcsinx)= 1/(1-x^2)^1/2
Note that if $\displaystyle \frac{\,d}{\,dx}\left[\arcsin\!\left(u\right)\right]=\frac{1}{\sqrt{1-u^2}}\cdot\frac{\,du}{\,dx}$, then it follows that $\displaystyle \frac{\,d}{\,dx}\left[\frac{1}{\arcsin\!\left(u\right)}\right]=\sqrt{1-u^2}\cdot\frac{\,dx}{\,du}$

Can you try this problem now?

also for x in domain [-1,1] prove that arcsinx + arccosx = pi/2
for this i took the derivatives of the LHS whiched equaled zero but dont know where to go from here

thanks
If you differntiate both sides you get $\displaystyle \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0\implies 0=0$, which would confirm the identity.

Does this make sense?

3. You could really substitute $\displaystyle \frac{\pi}{2}$ with any constant and that differentiation would hold true. I'll try to think of a way to show this algebraically, but if you think about it geometrically, it has to equal $\displaystyle \frac{\pi}{2}$ since those two operations are going to give you two angles of a right triangle which will add up to 90 degrees, or $\displaystyle \frac{\pi}{2}$.

4. thanks, im still a bit unsure about the first question

but for the second question that's what i got that it was equal to zero but i wasnt certain that was the answer

5. Originally Posted by b0mb3rz
thanks, im still a bit unsure about the first question
If $\displaystyle y=\frac{1}{\arcsin\!\left(x^2\right)}$, it follows that $\displaystyle y^{\prime}=\frac{1}{\frac{1}{\sqrt{1-\left(x^2\right)^2}}\cdot2x}=\boxed{\frac{\sqrt{1-x^4}}{2x}}$

Does this make sense?