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