Implicit Differentiation

So far, all the functions being differentiated are explicit functions, meaning that one of the variables was specifically given in terms of the other variable.

ex.

f(x) = 3x² + 1 then f '(x) = 6x

However, not all functions are given explicitly and are only implied by an equation.

ex.
xy = 1 is an equation given implicitly, explicitly it is y = 1/ x. Now to find dy/dx for xy = 1, simply solve for y and differentiate.

xy = 1
y = 1 / x = x^-1
dy/dx = -1 x^-2= 1/x²

But, not all equations are easily solved for y, as in the equation 3x + y³ = y² + 4, This is where implicit differentiation is applied. Implicit differentiation is taking the derivative of both sides of the equation with respect to one of the variables. Most commonly, used is the derivative of y with respect to x. or dy/dx. Since we have not solved for y as a function of x, the derivative of y must be left as dy/dx.

From the example above: 3x + y³ = y² + 4, solve for dy/dx.

3x + y³ = y² + 4
d/dx(3x + y³) = d/dx(y² + 4)
3 + 3y² dy/dx = 2y dy/dx

3 = 2y dy/dx - 3y² dy/dx
3 = y ( 2 - 3y ) dy/dx
3 / y (2 - 3y ) = dy/dx

Finding the slope of a curve implicitly:

ex.

find the slope of the curve x² + y³ = 2x + y at ( 2,4)

d/dx [x² + y³] = d/dx [2x + y]
2x + 3y² dy/dx = 2 + dy/dx
2x - 2 = (-3y² + 1) dy/dx
2( x - 1) / (-3y² + 1) = dy/dx = slope of the curve

substitute (2,4) into dy/dx to find the slope at that point.

2(2-1) / (-3 · 4² + 1) = 2 /-49 = -2/49 is the slope of the curve.