Implicit differentiation with partial derivatives

When an equation mixes its variables together, it can be hard to solve for one of them before differentiating. We can get the derivative directly by moving everything to one side and taking partial derivatives. Let’s do that first for two variables, then for three.

Write the equation as F equal to zero

If F(x,y)=0 and y is a differentiable function of x, the chain rule gives Fx+Fydydx=0. Solve for the derivative wherever Fy≠0.

dydx=-FxFy

Find the derivative when x and y are mixed together

Let’s say xe2y=x2-y3. Move everything to one side. When we take Fx, we hold y fixed. When we take Fy, we hold x fixed.

F(x,y)=x2-y3-xe2yFx=2x-e2yFy=-3y2-2xe2y
dydx=2x-e2y3y2+2xe2y

Find the two partial derivatives of z

For F(x,y,z)=0 with z=z(x,y), the same chain rule gives zx=-FxFz and zy=-FyFz wherever Fz≠0. To find zx, hold y fixed. To find zy, hold x fixed.

For x2sinz=3yz+2x3, take the partial derivatives of F first, treating x, y, and z as independent inputs.

F(x,y,z)=3yz+2x3-x2sinzFx=6x2-2xsinzFy=3zFz=3y-x2cosz
zx=-6x2-2xsinz3y-x2coszzy=-3z3y-x2cosz