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

Question

Emilia Aguilar · Accepted Answer

#-(y^2-ysec^2(x/y))/(xsec^2(x/y)+y^2)# #"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)#

Evelyn Agard · Answer

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

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)#

Elijah Abram · Answer

undefined To find the derivative of $ \tan\left(\frac{x}{y}\right) = x + y $, we can use implicit differentiation:

$$ \frac{d}{dx} \left[ \tan\left(\frac{x}{y}\right) \right] = \frac{d}{dx} (x + y) $$

Using the chain rule for the left side and the sum rule for the right side, we get:

$$ \sec^2\left(\frac{x}{y}\right) \cdot \frac{1}{y} = 1 $$

Now, solve for $ \frac{dy}{dx} $:

$$ \frac{dy}{dx} = y \sec^2\left(\frac{x}{y}\right) $$