Derivatives of arctan((x+y)/(1-xy))

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In summary, to find the second partial derivatives of z=arctan((x+y)/(1-xy)), you can use the addition formula for tangent to simplify the function and then use the derivative of arctan to find the first derivative. To make it easier, you can also use math notations by clicking on the "advanced" tab and using the Σ symbol.
  • #1
newyorkcity
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Homework Statement



Find all second partial derivatives of
z=arctan((x+y)/(1-xy))

Homework Equations



d/dx of arctan(x) is 1/(1+x^2)

The Attempt at a Solution



Not sure how to proceed... I don't want the answer, just an idea as to how to move forward.

My attempt at finding the first derivative...

z'=(1/(1+(x+y/(1-xy)) * (x(1-xy) - (x+y(-y)) / (1-xy)^2

Is this correct? If it is, I honestly don't know how to find the second derivative...

On another note, can someone tell me how I can use math notation instead of plain text, to make the equations and such easier to read? Thanks guys.
 
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  • #2
Take care of the parentheses. Check the derivative of the arctan function.

To make your task easier, remember the addition formula :
[tex]tan(\alpha+\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha)tan(\beta)}[/tex]
Can you write the function z(x,y) in a simpler form?
As for Math notations, go to "advanced" and click on the 'Σ'.
ehild
 

Related to Derivatives of arctan((x+y)/(1-xy))

What is the formula for finding the derivative of arctan((x+y)/(1-xy))?

The formula for finding the derivative of arctan((x+y)/(1-xy)) is:
d/dx [arctan((x+y)/(1-xy))] = (1 + (x+y)^2) / (1 + ((x+y)/(1-xy))^2)

What is the domain of arctan((x+y)/(1-xy))?

The domain of arctan((x+y)/(1-xy)) is all real numbers except for values that make the denominator equal to 0. In this case, the denominator is equal to 1-xy, so the domain would be all real numbers except for x=1 and y=1.

How do you simplify arctan((x+y)/(1-xy))?

To simplify arctan((x+y)/(1-xy)), you can use the trigonometric identity: arctan(u/v) = arctan(u) + arctan(v). This would give you the simplified form of arctan(x) + arctan(y).

Can you use the chain rule to find the derivative of arctan((x+y)/(1-xy))?

Yes, you can use the chain rule to find the derivative of arctan((x+y)/(1-xy)). You would need to substitute u = (x+y)/(1-xy) and use the chain rule formula: d/dx [f(g(x))] = f'(g(x)) * g'(x).

What is the graph of the derivative of arctan((x+y)/(1-xy))?

The graph of the derivative of arctan((x+y)/(1-xy)) would be a smooth curve with a maximum value of 1 and a minimum value of -1. It would also have vertical asymptotes at x=1 and y=1.

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