Question

√((x^(2)+y^(2))=tan((π/4)(x+y)). I tried solving using implicit differentiation and got (π-1)/(4-π). I don't think this is right. Can anyone help?

Answer 1

\[\sqrt{x^2+y^2}=\tan(\frac{\pi}{4}(x+y))\]?

Answer 2

and I guess you are suppose to find y'?

Answer 3

yes

Answer 4

\[x^2+y^2=(\tan(\frac{\pi}{4}(x+y))^2 \]

Answer 5

is that right though
it looks like you found y' evaluated at a certain point of (x,y)

Answer 6

Well ok I guess I will help with finding y' then

Answer 7

so I guess you didn't have a problem differentiating the x^2 and y^2 with w.r.t. x?

Answer 8

\[2x+2yy'=\frac{d (\tan(\frac{\pi}{4}(x+y))^2}{dx}\]
The other side requires chain rule

Answer 9

what was your stab at differentiating that part...

Answer 10

I'm going to do it it again. I went about it a little differently, so I'll do it your way

Answer 11

Ok I'm around a little longer but gonna grab me some cheese toast.

Answer 12

is the right side 2tan(π/4(x+y)((π/4)(π/4)(dy/dx))

Answer 13

I'm still struggling with it. It doesn't seem right.

Answer 14

well i don't see that you differentiated the tan part

Answer 15

starting from outside:
d(x^2)/dx=2x
d(tan(x))/dx=sec^2(x)
d(pi/4(x+y))=pi/4+pi*y'/4

Answer 16

\[\frac{d}{dx}([\tan(\frac{\pi}{4}(x+y))^2)=2[\tan(\frac{\pi}{4}(x+y))]^{2-1}\cdot \sec^2(\frac{\pi}{4}(x+y))\cdot \frac{\pi}{4}(1+y')\]