What is the derivative of arctan (xy) with respect to x?

[snoggerT]

whats derivative of arctan (xy) with respect to x

The Attempt at a Solution

- I feel kind of stupid for this one, but I can't seem to figure out the derivative here. I know with one variable that arctan goes to 1/(1+x^2), but I'm not sure what to do with two variables. please help.

 

[konthelion]

Since you know that \frac{d}{dx} \arctan x = \frac{1}{1+x^2}, you have to apply the chain rule.

 

[Defennder]

The more general formulae would be \frac{d}{dx} \arctan f(x) = \frac{f'(x)}{1+f(x)^2}. You have to use implicit differentiation and the chain rule to do this one.

 

[snoggerT]

The more general formulae would be \frac{d}{dx} \arctan f(x) = \frac{f'(x)}{1+f(x)^2}. You have to use implicit differentiation and the chain rule to do this one.

- I'm still not grasping it. We haven't really done any problems with implicit differentiation like this.

 

[rocomath]

y'=\frac{dy}{dx}

y=\tan^{-1}{(xy)} \rightarrow y=\tan^{-1}f(x)

Let f(x)=xy \rightarrow f'(x)=xy'+y

y'=\frac{f'(x)}{1+[f(x)]^2}

Now you take it.

 

[rootX]

try finding derv of
arctan (5x)
or
arctan(10x)
..
or
arctan(a.x)
..
now substitute your y for a

 

[tiny-tim]

Or d(arctan(xy))/dx = d(arctan(xy))/d(xy) times (d(xy)/dx).