How do you implicitly differentiate #-1=xy+e^ysec(x/y) #?

Question

How do you implicitly differentiate #-1=xy+e^ysec(x/y)  #?

Charles Anctil · Accepted Answer

I got

#(dy)/(dx) = (-y - 1/ye^ysec(x/y)tan(x/y))/[x - e^ysec(x/y)(x/y^2tan(x/y) + 1)]#

Since #y = f(x)#, we must account for that when we take the derivative of #y# with respect to #x#.

#d/(dx)[-1] = d/(dx)[xy + e^ysec(x/y)]#

#0 = (x(dy)/(dx) + y) + (e^yd/(dx)[sec(x/y)] + sec(x/y)d/(dx)[e^y])#

#= (x(dy)/(dx) + y) + (e^ysec(x/y)tan(x/y)*d/(dx)[x/y] + sec(x/y)e^y(dy)/(dx))#

#= x(dy)/(dx) + y + [e^ysec(x/y)tan(x/y)*(-x/y^2(dy)/(dx)+1/y) + sec(x/y)e^y(dy)/(dx)]#

Now we simplify and isolate #(dy)/(dx)#.

#0 = x(dy)/(dx) + y + [-x/y^2e^ysec(x/y)tan(x/y)(dy)/(dx) + 1/ye^ysec(x/y)tan(x/y) + sec(x/y)e^y(dy)/(dx)]#

#0 = x(dy)/(dx) + y - x/y^2e^ysec(x/y)tan(x/y)(dy)/(dx) + 1/ye^ysec(x/y)tan(x/y) + sec(x/y)e^y(dy)/(dx)#

It's all multiplied out, so separate your variables.

#- y - 1/ye^ysec(x/y)tan(x/y) = x(dy)/(dx) - x/y^2e^ysec(x/y)tan(x/y)(dy)/(dx) + sec(x/y)e^y(dy)/(dx)#

#- y - 1/ye^ysec(x/y)tan(x/y) = [x - x/y^2e^ysec(x/y)tan(x/y) + sec(x/y)e^y](dy)/(dx)#

#(dy)/(dx) = (-y - 1/ye^ysec(x/y)tan(x/y))/[x - x/y^2e^ysec(x/y)tan(x/y) + sec(x/y)e^y]#

Simplifying this further, we'd get:

#color(blue)((dy)/(dx) = (-y - 1/ye^ysec(x/y)tan(x/y))/[x - e^ysec(x/y)(x/y^2tan(x/y) + 1)])#

Emilia Aguilar · Answer

undefined To implicitly differentiate the equation $-1 = xy + e^y \sec(x/y)$, follow these steps:

1. Differentiate each term of the equation with respect to $x$.
2. Use the product rule for the term $xy$.
3. Apply the chain rule for the term $e^y \sec(x/y)$.

After differentiation, solve for $\frac{{dy}}{{dx}}$. The result is:

$$ \frac{{dy}}{{dx}} = \frac{{y - x \sec(x/y) \tan(x/y)}}{{x + ye^y \sec(x/y)}} $$