# Derivatives of Implicit Functions – Continuity and Differentiability | Class 12 Maths

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. Implicit Differentiation helps us compute the derivative of y with respect to x without solving the given equation for y, this can be achieved by using the chain rule which helps us express y as a function of x.

Implicit Differentiation can also be used to calculate the slope of a curve, as we cannot follow the direct procedure of differentiating the function y = f(x) and putting the value of the x-coordinate of the point in dy/dx to get the slope. Instead, we will have to follow the process of implicit differentiation and solve for dy/dx.

The method of implicit differentiation used here is a general technique to find the derivatives of unknown quantities.

**Example**

x

^{2}y^{2 }+ xy^{2}+ e^{xy}= abc = constant

**The function above is an implicit function, we cannot express x in terms of y or y in terms of x.**

**Derivative of Implicit Functions**

Since the functions can not be expressed in terms of one specific variable, we have to follow a different method to find the derivative of the implicit function :

While computing the derivative of the Implicit function, our aim is to solve for **dy/dx** or any higher-order derivatives depending on the function. To solve dy/dx in terms of x and y, we have to follow certain steps:

** Steps to compute the derivative of an implicit function**

- 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. For instance,

x

^{2}+ y^{2}= r^{2 (}Implicit function)

Differentiate with respect to x:

d(x^{2}) /dx + d(y^{2})/ dx = d(r^{2}) / dxSolve each term:

Using Power Rule: d(x^{2}) / dx = 2x

Using Chain Rule: d(y^{2})/ dx = 2y dydxr

^{2}is a constant, so its derivative is 0: d(r^{2})/ dx = 0Which 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

## Sample Problems on Derivative of Implicit Function

**Example 1. Find the expression for the first derivative of the function y(x) given implicitly by the equation: x ^{2}y^{3} – 4y + 3x^{3} = 2.**

**Solution:**

**Step 1: Differentiate the given equation or function with respect to x.**

x

^{2}y^{3}– 4y + 3x^{3}= 2.

d(x^{2}y^{3}– 4y + 3x^{3}) / dx = d(2) / dx

**Step 2: The right-hand side’s derivative will simply be 0 since it is a constant.**

The left-hand side after differentiation :

2xy

^{3}+3x^{2}y^{2 }* dy/dx – 4 * dy/dx + 9x^{2}= 0.

**Step 3: Collect the terms involving dy/dx on one side and take the remaining terms on the other side to get:**

dy/dx * (3x

^{2}y^{2}– 4) = -9x^{2}– 2xy^{3}dy/dx = – ( 9x

^{2}+ 2xy^{3}) / (3x^{2}y^{2}– 4 )

**This is the expression for the first derivative at any point on the curve. This expression also helps us compute the slope of a tangent drawn at point (x, y) to the curve.**

**Example 2. Find the first derivative of y, given implicitly as: y – tan ^{-1}y = x.**

**Solution:**

d(y – tan

^{-1}y) /dx = d(x)/ dx.dy/dx – (1/(1 + y

^{2}) * dy/dx = 1dy/dx =1/(1 / (1–1/ (1 + y

^{2})))dy/dx = 1/y

^{2}+ 1

**Example 3. Find dy/dx if x ^{2}y^{3} − xy = 10.**

**Solution:**

2xy

^{3}+ x^{2}. 3y^{2 }. dy/dx – y – x . dy/dx = 0(3x

^{2}y^{2}– x ) . dy/dx = y – 2xy^{3}dy/dx = (y – 2xy

^{3}) / (3x^{2}y^{2 }– x)

**Example 4. Find dy/dx if y = sinx + cosy **

**Solution:**

y – cosy = sinx

dy/dx + siny. dy/dx = cosx

dy/dx = cosx / (1 + siny)

**Example 5. Find the slope of the tangent line to the curve x ^{2}+ y^{2}= 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).

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