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

##### 1 Answer

#### Explanation:

When we differentiate

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

# 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

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

So Let

# (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