Differential of a acrtan

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MrJoe2000

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Nov 1, 2011
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I have this really heavy equation and I'm stuck on how to start my differentiation. I have a function:
​

alpha= tan^-1[y/(r-h)]​
What I need to now do is take the derivative with respect to psi. What I need to find is dAlpha/dPsi.

I know that the derivative of tan^-1= 1/(1+x^2) but does my "x" remain y/(r-h) or do I then have to differentiate that too? I guess what I'm asking is, is my dAlpha/dPsi= 1/(1+ (y/(r-h)) or is it something different?

Aaany help here would be really appreciated.
 
By psi, I reckon you mean \(\displaystyle \psi\)?.

What is psi?. I do not see it in your function. Just y,r, and h.

Or does \(\displaystyle \psi=\frac{y}{r-h}\)
 
By psi, i mean pressure. so I want to find differentiate with respect to a pressure. I think I should wind up with dy/dpsi and dh/dpsi.
 
MrJoe2000 said:
I have this really heavy equation and I'm stuck on how to start my differentiation. I have a function:

alpha= tan^-1[y/(r-h)]​
What I need to now do is take the derivative with respect to psi. What I need to find is dAlpha/dPsi.

I know that the derivative of tan^-1= 1/(1+x^2) but does my "x" remain y/(r-h) or do I then have to differentiate that too? I guess what I'm asking is, is my dAlpha/dPsi= 1/(1+ (y/(r-h)) or is it something different?

Aaany help here would be really appreciated.
Click to expand...

Let's define

u(p) = y(p)/[r-h(p)]..................... I assume r = constant

d/dp u = {y' * (r-h) + h' * y}/(r-h)2

then

d/du (α) = 1/(1+ u2)

d/dp (α) = d/du (α) * du/dp

and continue.....
 
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