Question

How do you find the derivative of `tan(x/y)=x+y`?

Answer 1

`-(y^2-ysec^2(x/y))/(xsec^2(x/y)+y^2)`

Explanation:

`"differentiate "tan(x/y)" using the "color(blue)"chain rule"`

`rArrd/dx(tan(x/y))`

`=sec^2(x/y)xxd/dx(x/y)`

`"differentiate "x/y" using the "color(blue)"quotient rule"`

`=sec^2(x/y)xx(y-x.dy/dx)/y^2`

`=(ysec^2(x/y)-xsec^2(x/y)dy/dx)/y^2`

`"returning to the original"`

`(ysec^2(x/y)-xsec^2(x/y)dy/dx)/y^2=1+dy/dx`

`rArrysec^2(x/y)-xsec^2(x/y)dy/dx=y^2+y^2dy/dx`

`rArrdy/dx(-xsec^2(x/y)-y^2)=y^2-ysec^2(x/y)`

`rArrdy/dx=-(y^2-ysec^2(x/y))/(xsec^2(x/y)+y^2)`

Answer 2

`dy/dx=(y(sec^2(x/y)-y))/(xsec^2(x/y)+y^2)`

Explanation:

Use implicit differentiation:

`d/dx(tan(x/y))=d/dx(x+y)`

You need the chain rule on the tangent part:

`sec^2(x/y)\*(y*(1)-x(dy/dx))/(y^2)=1+dy/dx`

Distribute on the left side:

`sec^2(x/y)/y-(xsec^2(x/y))/y^2*dy/dx=1+dy/dx`

Move everything with a `dy/dx` to the left and everything without to the right:

`-(xsec^2(x/y))/y^2*dy/dx-dy/dx=1-sec^2(x/y)/y`

Factor out the `dy/dx`:

`dy/dx*(-(xsec^2(x/y))/y^2-1)=1-sec^2(x/y)/y`

Get `dy/dx` by itself by dividing both sides by that coefficient:

`dy/dx=(1-sec^2(x/y)/y)/(-(xsec^2(x/y))/y^2-1)`

Get a common denominator in both the numerator and the denominator:

`dy/dx=((y-sec^2(x/y))/y)/((-xsec^2(x/y)-y^2)/y^2)`

Simplify the complex fraction (I think of it as keep, change, turn):

`dy/dx=(y^2-ysec^2(x/y))/(-xsec^2(x/y)-y^2)`

I factored the negative out of the denominator and distributed it to the numerator:

`dy/dx=(ysec^2(x/y)-y^2)/(xsec^2(x/y)+y^2)`

I decided to factor the common `y` out of the numerator:

`dy/dx=(y(sec^2(x/y)-y))/(xsec^2(x/y)+y^2)`