# Thread: i need some help deriving inverse trig functions

1. ## i need some help deriving inverse trig functions

i have to show that dy/dx= (x+y)/(x-y) if arctan (y/x)= log(sq.rt.(x^2 + y^2))

i was able to get everything before this, but i dont really understand what the questions asking. also, i just did a whole page of tanh x, so that keeps interfering with my memory of arctan.

any help would be very appreciative, thanks alot!

2. Originally Posted by twostep08
i have to show that dy/dx= (x+y)/(x-y) if arctan (y/x)= log(sq.rt.(x^2 + y^2))

i was able to get everything before this, but i dont really understand what the questions asking. also, i just did a whole page of tanh x, so that keeps interfering with my memory of arctan.

any help would be very appreciative, thanks alot!
You said that you just did a whole page of $\displaystyle tanh(x)$??? What does the hyperbolic tangent have to do with this?

This is the expression:

$\displaystyle Arctan\left(\frac{y}{x}\right)=log(\sqrt{x^2+y^2})$

All you need to do is use implicit differentiation, and solve for $\displaystyle \frac{dy}{dx}$.

It's easy to show that:

$\displaystyle \frac{d}{dx}Arctan(x)=\frac{1}{1+x^2}$.

Use this to find the derivative of the left side:

$\displaystyle \frac{d}{dx}Arctan\left(\frac{y}{x}\right)=\frac{d }{du}Arctan(u)\frac{du}{dx}$ where $\displaystyle u=\frac{y}{x}$. You should be able to figure that out using implicit differentiation.

Use a similar argument to differentiate the implicit logarithmic function.

3. i was just using that to say that every time i see arctan i wanna throw in a load of e's. this is the first time i really learned either of them and i keep mixing them up, so i want to make sure i understand this properly before i learn it the wrong way

4. im sure that im doing this wrong, but heres what im getting on the left side..
u=y/x
du/dx=(xy'-y)/x^2

d/du arctan u du/dx= (1/(1+(y/x)^2))*[(xy'-y)/x^2]

and that looks pretty wrong, so i wanted to find out my misstep. also if im right, should i leave that alone before it do the right side for now or simplify it?

5. Originally Posted by twostep08
im sure that im doing this wrong, but heres what im getting on the left side..
u=y/x
du/dx=(xy'-y)/x^2

d/du arctan u du/dx= (1/(1+(y/x)^2))*[(xy'-y)/x^2]

and that looks pretty wrong, so i wanted to find out my misstep. also if im right, should i leave that alone before it do the right side for now or simplify it?
Yes, that's the correct derivative. All you should do is multiply through out. You should get $\displaystyle \frac{xy'-y}{x^2+y^2}$. All I did was mulitply. So now you just have to solve:

$\displaystyle \frac{xy'-y}{x^2+y^2}=\frac{d}{dx}log(\sqrt{x^2+y^2})$

$\displaystyle \frac{d}{dx}log(\sqrt{x^2+y^2})=\frac{1}{\sqrt{x^2 +y^2}}\frac{d}{dx}\sqrt{x^2+y^2}$

$\displaystyle =\frac{1}{\sqrt{x^2+y^2}}\left(\frac{2x+2yy'}{2\sq rt{x^2+y^2}}\right)$

$\displaystyle =\frac{x+yy'}{x^2+y^2}$

Now we have: $\displaystyle \frac{xy'-y}{x^2+y^2}=\frac{x+yy'}{x^2+y^2}$

This gives $\displaystyle y'=\frac{x+y}{x-y}$