What is the slope of the tangent line of # (x+2y)^2/(1-e^y) =C #, where C is an arbitrary constant, at #(1,1)#?

Question

1283 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-04T21:18:10

#dy/dx = (2e-2)/(4-e)#

First find the derivative of the function and then plug in 1 for x and 1 for y
#(x+2y)^2/(1-e^y)=C#

Let's find the derivative of the function with respect to x and hold y constant.

#f_x =2(x+2y)*1/(1-e^y)=( 2(x+2y))/(1-e^y)#

Then find the derivative of the function with respect to y and hold x constant

Use the product rule #f_y=(gf'-fg')/g^2#

#f_y=((1-e^y)*4(x+2y)-(x+2y)^2(-e^y))/(1-e^y)^2#

#f_y=((x+2y)[4(1-e^y)+e^y(x+2y)])/(1-e^y)^2#

Now to find the slope #dy/dx# use the formula #dy/dx=-f_x/f_y#

#dy/dx=-(f_x/f_y) =-( ( 2(x+2y))/(1-e^y))/(((x+2y)[4(1-e^y)+e^y(x+2y)])/(1-e^y)^2)#

#dy/dx=-(f_x/f_y) =-( ( 2(cancel (x+2y)))/cancel(1-e^y)) *((1-e^y)^cancel 2)/(cancel ((x+2y))[4(1-e^y)+e^y(x+2y)])#

#dy/dx=-(f_x/f_y) =(-2+2e^y)/(4-4e^y+xe^y+2ye^y)#

#dy/dx=-(f_x/f_y) =(-2+2e^1)/(4-4e^1+(1)e^1+2(1)e^1)#

#dy/dx = (2e-2)/(4-e)#