Prove Quotient Rule formula Using Implicit Differentiation

One way to obtain an expression’s derivative or differentiation in calculus is to take the ratio or division of two differentiable functions and divide it by the expression. By this, I mean that while trying to get the derivative of f(x)/g(x), we can use the rule of quotients, because both of these functions are differentiable and g(x) is equal to zero. Products and derivation limitations in differentiation are closely linked to the rule’s application. In the following sections, we’ll go over the formula for the quotient rule and its proof using solved instances.

Proving Quotient Rule Formula using Implicit Differentiation : 

The quotient formula was discussed in depth in the previous chapter. This formula was used to get functions that can be divided by a differentiable quotient which is what we learned about in this section. In the following section, we will look at how to display the quotient rule formula. There are a variety of methods for demonstrating the quotient rule formula, including, but not limited to, the following examples:

Using limit and derivative properties

Differentiating implicitly

Applying the chain rule

Differentiating implicitly : 

To prove the quotient rule formula using the implicit differentiation formula, let’s start with a differentiable function f(x) = u(x)/v(x), so u(x) = f(x)v (x). Using the product rule, we get:

 u'(x) = f'(x)v(x) + f(x)v’ (x). When we figure out f'(x), we get,

Implicit differentiation : 

Even if such a function exists, some equations in x and y do not clearly define y as a function x and cannot be easily modified to solve for y in terms of x. When this happens, it’s proposed that there’s a function y = f(x) that can satisfy the given equation. You can acquire the derivative of y with respect to x via implicit differentiation without having to solve the supplied equation for y. The chain rule must be employed anytime the function y is being differentiated because of our assumption that y may be represented as a function of x.

We cannot start with dy/dx directly in the process of implicit differentiation because an implicit function is not of the form y = f(x), but rather of the form f(x, y) = 0. Therefore, we cannot start with dy/dx directly. Before mastering the method of implicit differentiation, we need to be familiar with the derivative rules, such as the power rule, product rule, quotient rule, chain rule, and so on. The following is a flowchart illustrating the stages involved in executing implicit differentiation.

Rules of Implicit Differentiation :

These processes will now be demonstrated with an example in which the implicit derivative dy/dx will be obtained if the function is y + sin y = sin x and the function would be y + sin y = sin x.

1. Differentiate every term on both sides of the equation with respect to x (Step 1).

As a result, we have d/dx(y) + d/dx(sin y) = d/dx(y) (sin x).

Step 2: Use the derivative formulas to get the derivatives, as well as the chain rule, to complete the computation.

The derivative formulas should be used to directly differentiate all of the x terms; however, when differentiating the y terms, multiply the actual derivative by the derivative factor dy/dx.

In this case, d/dx (sin x) equals cos x, but d/dx (sin y) equals cos y (dy/dx) in this case.

Then the preceding process is transformed into:

(dy/dx) + (cos y) (dy/dx) = cos x = (dy/dx) + (cos y) = cos x

Step 3: Find the solution for dy/dx.

Assuming that dy/dx is a common factor:

The ratio (dy/dx) is (1 + cos y) = cos x. dy/dx is equal to (cos x)/(1 + cos y).

This is referred to as the implicit derivative.

Conclusion : 

One way to obtain an expression’s derivative or differentiation in calculus is to take the ratio or division of two differentiable functions and divide it by the expression. By this, I mean that while trying to get the derivative of f(x)/g(x), we can use the rule of quotients, because both of these functions are differentiable and g(x) is equal to zero. Products and derivation limitations in differentiation are closely linked to the rule’s application. Even if such a function exists, some equations in x and y do not clearly define y as a function x and cannot be easily modified to solve for y in terms of x. When this happens, it’s proposed that there’s a function y = f(x) that can satisfy the given equation.