[Maths Class Notes] on Implicit Function Differentiation Pdf for Exam

Differentiation is one of the building blocks of calculus. It calculates the rate of change of a given quantity. Today, we use differentiation in almost every aspect of physics, mathematics, and chemistry. Do you want to calculate the acceleration of a sprinter exactly after 5 seconds of start? Or do you want to calculate the displacement of a moving body traveling at a certain speed in the given time? You can apply differential calculus in various real-time problems and find the correct solution. 

Differential calculus is the area of study that calculates the derivatives of a function, and the process of calculating derivatives is known as differentiation. For a given function y=f(x), the derivative of y is (dy/dx) where dy represents the change in y and dx represents the change in x. 

                                            

(tangent on a function y=f(x))

In the above graph, we can calculate the derivative of the given function y=f(x)by determining the slope of the tangent line P. We mainly use differentiation on two types of functions: explicit function and implicit function. So, you must be wondering what is the Implicit Function?

Implicit Function

Implicit Function definition conveys when we are not able to isolate the dependent variable in an equation that function becomes an implicit function. Both the dependent variables and independent variables are present in this type of function. For eg: x² +y²= 16 . Whereas explicit function is present in terms of the independent variable. Consider y=f(x) where x is an independent variable, and y is a dependent variable.

In explicit function, y is in terms of x. According to implicit function meaning, a function is expressed in both y and x. 

Differentiation Of Implicit Functions

We can easily differentiate explicit functions using chain rule or rule of differentiation. But sometimes, it can be challenging to express an equation explicitly. Suppose if we were to reduce the equation x² +y²= 16 in terms of x them we get 

Y = + [sqrt{(16 – x ^{2}}] and y = – [sqrt{(16 – x ^{2}}] . We perform implicit differentiation in such cases. 

The implicit differentiation meaning isn’t exactly different from normal differentiation. Since we cannot reduce implicit functions explicitly in terms of independent variables, we will modify the chain rule to perform differentiation without rearranging the equation. In the above example, we will differentiate each term in turn, so the derivative of y² will be 2y*dy/dx. Let us look at implicit differentiation examples to understand the concept better.

Solved Examples

Example 1: 

What is implicit function differentiation of x² +y²= 16?

Answer: 

Using implicit function definition

We will perform Differentiation of implicit functions on both sides and each term w.r.t x.

2x + 2y(dy/dx)=0

Rearranging the equation

2y(dy/dx)=-2x

Solving the equation

dy/dx= -2x/2y

Derivative of implicit function is dy/dx= -x/y

Let us look at some other examples.

Example 2: 

Find dy/dx If y=sin(x) + cos(y)

Answer:

According to implicit function meaning the given function is implicit. 

Hence, we will calculate the derivative of implicit function without rearranging the equation.

Performing Differentiation of implicit functions on both sides and each terms with respect to x.

dy/dx=cos(x)-sin(y)*dy/dx

Rearranging the above equation

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

dy/dx(1+sin(y))=cos(x)

Solve the equation

dy/dx=cos(x)/1+sin(y) 

Example 3:  

Differentiate 10x⁴ – 18xy² + 10y³ = 48 with respect to x.

Answer:

The implicit function meaning holds true for the given function.

Hence, we will use the product rule of differentiation on xy² i.e (FG)’ = F G’ + F’ G

Let us find out the derivative of Implicit function by differentiating each term in the equation w.r.t x.

10 (4x²) – 18(x(2y * dydx) + y²) + 10(3y² * dydx) = 0

Let us further simplify the above equation.

40x³ – 36xy * dydx – 18y² + 30y² * dydx = 0

Now we bring all dy/dx on the left side and rest of the terms on the right side

-36xy * dydx + 30y² * dydx = 40x³ + 18y²

Taking dy/dx common

(30y² – 36xy) dydx = 18y² – 40x²

Divide both sides by two and solve the equation

dy/dx = 9y² – 20x²/(15y² – 18xy)