4. Derivatives of implicit functions

Derivatives of implicit functions:

Implicit functions are functions where a specific variable cannot be expressed as a function of the other variable. A function that depends on more than one variable.

 we’ll adopt the following procedure:

• Given an implicit function with the dependent variable y and the independent variable x (or the other way around).
• Differentiate the entire equation with respect to the independent variable (it could be x or y).
• After differentiating, we need to apply the chain rule of differentiation.
• Solve the resultant equation for dy/dx (or dx/dy likewise) or differentiate again if the higher-order derivatives are needed.

 “Some function of y and x equals to something else”. Knowing x does not help us compute y directly.

Example, x² + y² = r^{2 (}Implicit function) 

Differentiate with respect to x:
d(x²) /dx + d(y²)/ dx = d(r²) / dx 

Solve each term:

Using  Power Rule: d(x²) / dx = 2x

Using Chain Rule :  d(y²)/ dx = 2y dydx

r² is a constant, so its derivative is 0:  d(r²)/ dx  = 0

Which gives us:

2x + 2y dy/dx = 0

Collect all the  dy/dx on one side

y dy/dx = −x

Solve for dy/dx:

dy/dx = −xy     

Example . Find dy/dx if x²y³ − xy = 10.

Solution:

2xy³ + x². 3y² . dy/dx – y – x . dy/dx = 0

(3x²y² – x ) . dy/dx = y – 2xy³

dy/dx = (y – 2xy³) / (3x²y² – x) 

Example . Find dy/dx if y = sinx + cosy 

Solution:

y – cosy = sinx

dy/dx + siny. dy/dx = cosx

dy/dx = cosx / (1 + siny)

Example . Find the slope of the tangent line to the curve x²+ y²= 25 at the point (3,−4).

Solution:

Note that the slope of the tangent line to a curve is the derivative, differentiate implicitly with respect to x, which yields,

2x + 2y. dy/dx = 0

dy/dx = -x/y

Hence, at (3,−4), y′ = −3/−4 = 3/4, and the tangent line has slope 3/4 at the point (3,−4).