# i need some help deriving inverse trig functions

• Dec 3rd 2009, 11:21 AM
twostep08
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!
• Dec 3rd 2009, 11:41 AM
Quote:

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.
• Dec 3rd 2009, 11:54 AM
twostep08
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
• Dec 3rd 2009, 12:12 PM
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?
• Dec 3rd 2009, 01:16 PM
Quote:

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