Implicit Differentiation

Implicit Differentiation

Hey there, JEE aspirants! Implicit differentiation might sound intimidating, but trust me, it's a powerful technique that pops up frequently in JEE Main. It's all about finding derivatives when you can't easily express $y$ directly in terms of $x$. So, let's dive in and master this essential tool!

1. Implicit vs Explicit Functions

Before we get into the nitty-gritty, let's clarify the difference between explicit and implicit functions. This is crucial for understanding when to use implicit differentiation.

Explicit Function: An explicit function is one where $y$ is clearly isolated on one side of the equation, expressed directly in terms of $x$. For example:

• $$y = x^2 + 3x - 5$$
• $$y = \sin(x) + e^x$$

Here, you can directly substitute a value for $x$ and calculate the corresponding value of $y$.

Implicit Function: An implicit function, on the other hand, is defined by an equation where $y$ is not isolated. You can't easily (or at all!) rewrite the equation to get $y = f(x)$. For example:

• $$x^2 + y^2 = 25$$ (Equation of a circle)
• $$x^3 + xy + y^3 = 7$$
• $$\sin(xy) + x^2 = y$$

In these cases, $x$ and $y$ are intertwined, and it's difficult or impossible to solve for $y$ explicitly. While $x^2 + y^2 = 25$ *could* be solved for $y$, it would require splitting it into two functions ($y = \sqrt{25-x^2}$ and $y = -\sqrt{25-x^2}$), which is cumbersome. That’s where implicit differentiation comes in handy.

2. Differentiating Implicitly Defined Functions

The core idea behind implicit differentiation is to differentiate both sides of the equation with respect to $x$, remembering that $y$ is a function of $x$. This is where the chain rule becomes essential. Let's break it down:

The Chain Rule Connection: Since $y$ is a function of $x$, whenever we encounter a term involving $y$, we need to apply the chain rule. Remember the chain rule: $d/dx[f(g(x))] = f'(g(x)) \cdot g'(x)$. In our case, $g(x) = y$, so $g'(x) = dy/dx$.

Let's illustrate with an example: Consider the equation of a circle, $$x^2 + y^2 = 25$$.

1. Differentiate both sides with respect to $x$: $$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$$
2. Apply the sum rule: $$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$
3. Differentiate each term: $$2x + 2y \frac{dy}{dx} = 0$$

   Notice how we used the chain rule to differentiate $y^2$. We treated $y$ as a function of $x$, so $d/dx[y^2] = 2y \cdot dy/dx$. The derivative of the constant 25 is 0.

3. Finding dy/dx from Equations

Now that we've differentiated implicitly, we can solve for $dy/dx$. Let's continue with our circle example, $$x^2 + y^2 = 25$$:

1. We had: $$2x + 2y \frac{dy}{dx} = 0$$
2. Isolate the term with $dy/dx$: $$2y \frac{dy}{dx} = -2x$$
3. Solve for $dy/dx$: $$\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}$$

So, for the circle $x^2 + y^2 = 25$, the derivative $dy/dx = -x/y$. This tells us the slope of the tangent line to the circle at any point $(x, y)$.

Let's look at another example: $$x^3 + xy + y^3 = 7$$

1. Differentiate both sides: $$\frac{d}{dx}(x^3 + xy + y^3) = \frac{d}{dx}(7)$$
2. Apply sum and product rules: $$\frac{d}{dx}(x^3) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^3) = 0$$ This becomes: $$3x^2 + (x\frac{dy}{dx} + y) + 3y^2\frac{dy}{dx} = 0$$
3. Isolate $dy/dx$ terms: $$x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = -3x^2 - y$$
4. Factor out $dy/dx$: $$\frac{dy}{dx}(x + 3y^2) = -3x^2 - y$$
5. Solve for $dy/dx$: $$\frac{dy}{dx} = \frac{-3x^2 - y}{x + 3y^2}$$

JEE-Specific Tricks (When Applicable)

Implicit differentiation problems in JEE Main often involve finding the equation of a tangent line at a specific point. Here's how to tackle those:

1. Find $dy/dx$ using implicit differentiation.
2. Substitute the given $(x, y)$ coordinates into the expression for $dy/dx$ to find the slope of the tangent line at that point.
3. Use the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the given point, to find the equation of the tangent line.

Example: Find the equation of the tangent line to the curve $x^2 + y^2 = 25$ at the point (3, 4).

We already found that $dy/dx = -x/y$. At the point (3, 4), $dy/dx = -3/4$. So, the slope of the tangent line is -3/4.

Using the point-slope form: $y - 4 = (-3/4)(x - 3)$. Simplifying, we get $y = (-3/4)x + 9/4 + 4$, or $y = (-3/4)x + 25/4$. Therefore, the equation of the tangent line is $3x + 4y = 25$.

Implicit differentiation is a valuable skill for JEE Main. Master the concepts, practice diligently, and you'll be well-prepared to tackle any related problems!