How do you implicitly differentiate #2=e^(xy)-ln(x^3y)+siny #?

1 Answer
Feb 26, 2017

# dy/dx = (3y-xy^2e^(xy))/(x^2ye^(xy) -x+xycosy) #

Explanation:

When we differentiate #y# wrt #x# we get #dy/dx#. However, we cannot differentiate a non implicit function of #y# wrt #x#. But if we apply the chain rule we can differentiate a function of #y# wrt #y# but we must also multiply the result by #dy/dx#.

When this is done in situ it is known as implicit differentiation.

We have:

# 2 = e^(xy) - ln(x^3y) + siny #

First let's simplify the expression using the rules for logarithms:

# \ \ \ \ \ e^(xy) - ln(x^3) - ln(y) + siny = 2#
# :. e^(xy) - 3ln(x) - ln(y) + siny = 2#

Differentiate wrt #x#:

# e^(xy)(y+xdy/dx) -3/x-1/ydy/dx+cosydy/dx = 0#

# ye^(xy) +xe^(xy)dy/dx -3/x-1/ydy/dx+cosydy/dx = 0#

# xy^2e^(xy) +x^2ye^(xy)dy/dx -3y-xdy/dx+xycosydy/dx = 0#

# x^2ye^(xy)dy/dx -xdy/dx +xycosydy/dx = 3y-xy^2e^(xy) #

# (x^2ye^(xy) -x+xycosy)dy/dx = 3y- xy^2e^(xy) #

# dy/dx = (3y-xy^2e^(xy))/(x^2ye^(xy) -x+xycosy) #

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find #y# explicitly as a function of #x#, only implicitly through the equation #F(x, y) = 0# which defines #y# as a function of #x, y = y(x)#. Therefore we can write #F(x, y) = 0# as #F(x, y(x)) = 0#. Differentiating both sides of this, using the partial chain rule gives us

# (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)) #

So Let # F(x,y) =e^(xy) - ln(x^3y) + siny-2#; Then:

# (partial F)/(partial x) = ye^(xy) - 3/x \ \ #; and #\ \ (partial F)/(partial x) = xe^(xy) - 1/y + cosy #

And so:

# dy/dx = -(ye^(xy) - 3/x)/(xe^(xy) - 1/y + cosy) #

# \ \ \ \ \ = -{1/x(xye^(xy) - 3)}/{1/y(xye^(xy) - 1 + ycosy)}#

# \ \ \ \ \ = -y/x{(xye^(xy) - 3)}/{(xye^(xy) - 1 + ycosy)}#

# \ \ \ \ \ = {3y-xy^2e^(xy )}/{(x^2ye^(xy) - x +xycosy)}#

Here we get an immediate implicit function for the derivative, so here if you are familiar with partial differentiation the solution is much easier