Step-by-Step Guide to Solving Implicit Differentiation Problems
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 = +
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)
FAQs on Implicit Function Differentiation Explained
1. What is an implicit function, and how does it differ from an explicit function?
An explicit function is one where the dependent variable (usually 'y') is expressed directly in terms of the independent variable (usually 'x'). For example, y = 3x² + 2 is an explicit function because 'y' is isolated on one side. In contrast, an implicit function is one where the relationship between 'x' and 'y' is mixed within the equation, and 'y' is not isolated. A classic example is the equation of a circle, x² + y² = 25. Solving for 'y' explicitly is often difficult or impossible.
2. What are the key steps to perform implicit differentiation?
Implicit differentiation is a straightforward process used to find the derivative dy/dx for an implicit function. The method involves these core steps as per the CBSE Class 12 syllabus:
Differentiate Both Sides: Differentiate every term in the equation with respect to 'x'.
Apply the Chain Rule: When differentiating a term that includes 'y', you must apply the chain rule. For instance, the derivative of y² with respect to x is 2y multiplied by dy/dx.
Isolate dy/dx: Use algebra to gather all terms containing dy/dx on one side of the equation and all other terms on the opposite side.
Solve for dy/dx: Factor out dy/dx and divide to find the final expression for the derivative.
3. Why is the chain rule so important when differentiating an implicit function?
The chain rule is the fundamental principle that makes implicit differentiation work. Since 'y' is considered a function of 'x' (even if we can't write it out explicitly), any operation on 'y' is an operation on a function. When we differentiate with respect to 'x', the chain rule dictates that we must first differentiate the outer function (e.g., the square in y²) and then multiply by the derivative of the inner function (which is dy/dx). Without applying the chain rule to every 'y' term, the resulting derivative would be incorrect as it would fail to account for the relationship between 'y' and 'x'.
4. In what situations is it necessary to use implicit differentiation?
Implicit differentiation is not just a choice; it's often a necessity. You should use it in situations where it is either very difficult or impossible to express 'y' as an explicit function of 'x'. This commonly occurs with equations defining geometric shapes like circles (x² + y² = r²), ellipses, or more complex curves like x³ + y³ = 3axy. It allows us to find the slope of the tangent line at any point on such a curve without needing to first solve for 'y'.
5. How is implicit differentiation used to find the derivatives of inverse trigonometric functions?
Implicit differentiation provides an elegant method to derive the formulas for derivatives of inverse trigonometric functions. For example, to find the derivative of y = sin⁻¹(x):
Rewrite the function: First, express it without the inverse notation: sin(y) = x.
Differentiate implicitly: Differentiate both sides with respect to x: cos(y) * (dy/dx) = 1.
Solve for dy/dx: Rearrange to get dy/dx = 1/cos(y).
Substitute back to x: Using the identity sin²(y) + cos²(y) = 1 and knowing sin(y) = x, we find that cos(y) = √(1 - x²). This gives the final derivative: dy/dx = 1 / √(1 - x²).
6. What are some common examples of implicit functions studied in the CBSE Maths syllabus?
In the Class 12 CBSE curriculum, students encounter several types of implicit functions. Some common examples include:
Equations of conic sections, like a circle x² + y² = a² or an ellipse x²/a² + y²/b² = 1.
Polynomial-style relationships like x³ + y³ = 3xy.
Equations involving trigonometric and algebraic terms, such as y + sin(y) = cos(x).
Functions where variables are part of products or quotients, like xy = x + y.
7. Can implicit differentiation be applied to functions involving exponential or logarithmic terms?
Yes, absolutely. The rules of implicit differentiation apply to any type of function, including exponential and logarithmic ones. For example, if you have an equation like eʸ = x + y, you can differentiate it implicitly. Differentiating with respect to x, you get eʸ(dy/dx) = 1 + dy/dx. You can then group the dy/dx terms and solve. Similarly, for an equation like ln(xy) = 5, you would use the properties of logarithms and then apply implicit differentiation and the chain rule as needed.
