Logarithmic and Parametric Differentiation

Welcome, JEE aspirants! This lesson dives into two powerful differentiation techniques: logarithmic and parametric differentiation. These are essential tools in your calculus arsenal, frequently appearing in JEE Main problems. Mastering them will significantly boost your problem-solving speed and accuracy. Let's get started!

1. Logarithmic Differentiation: Products, Quotients, and Functions Raised to Functions

Logarithmic differentiation shines when dealing with complex expressions involving products, quotients, and, most importantly, functions raised to other functions. The core idea is to simplify these expressions using logarithms before differentiating.

Why it Works: Logarithms transform products into sums, quotients into differences, and exponents into products. This allows us to break down complex differentiation problems into simpler ones.

Consider a function like: $y = \frac{u(x) \cdot v(x)}{w(x)}$

Differentiating this directly can be messy. Instead, let's use logarithmic differentiation:

Take the natural logarithm of both sides: $\ln y = \ln \left( \frac{u(x) \cdot v(x)}{w(x)} \right)$
Apply logarithm properties: $\ln y = \ln u(x) + \ln v(x) - \ln w(x)$
Differentiate both sides with respect to $x$ : $\frac{1}{y} \cdot \frac{dy}{dx} = \frac{u'(x)}{u(x)} + \frac{v'(x)}{v(x)} - \frac{w'(x)}{w(x)}$
Multiply both sides by $y$ to solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = y \cdot \left( \frac{u'(x)}{u(x)} + \frac{v'(x)}{v(x)} - \frac{w'(x)}{w(x)} \right)$
Substitute the original expression for $y$ : $\frac{dy}{dx} = \frac{u(x) \cdot v(x)}{w(x)} \cdot \left( \frac{u'(x)}{u(x)} + \frac{v'(x)}{v(x)} - \frac{w'(x)}{w(x)} \right)$

Notice how the complex quotient rule is elegantly avoided!

Example: Differentiate $y = \frac{x^2 \cdot \sin x}{\sqrt{x}}$

$\ln y = \ln(x^2) + \ln(\sin x) - \ln(\sqrt{x}) = 2\ln x + \ln(\sin x) - \frac{1}{2}\ln x$
$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x} + \frac{\cos x}{\sin x} - \frac{1}{2x}$
$\frac{dy}{dx} = \frac{x^2 \sin x}{\sqrt{x}} \left( \frac{2}{x} + \cot x - \frac{1}{2x} \right)$

2. Differentiating Functions of the Form $u(x)^{v(x)}$

This is where logarithmic differentiation truly shines. Direct differentiation is usually impossible. The key formula is:

For $y = u(x)^{v(x)}$ :

Take the natural logarithm: $\ln y = v(x) \cdot \ln u(x)$
Differentiate both sides: $\frac{1}{y} \cdot \frac{dy}{dx} = v'(x) \ln u(x) + v(x) \cdot \frac{u'(x)}{u(x)}$
Solve for $\frac{dy}{dx}$ : $\frac{dy}{dx} = y \cdot \left( v'(x) \ln u(x) + v(x) \cdot \frac{u'(x)}{u(x)} \right)$
Substitute the original expression for $y$ : $\frac{dy}{dx} = u(x)^{v(x)} \cdot \left( v'(x) \ln u(x) + v(x) \cdot \frac{u'(x)}{u(x)} \right)$

Intuition: By taking logarithms, we bring the exponent down as a multiplier, making the expression amenable to the product rule.

Example: Differentiate $y = x^x$

$\ln y = x \ln x$
$\frac{1}{y} \frac{dy}{dx} = (1) \ln x + x \cdot \frac{1}{x} = \ln x + 1$
$\frac{dy}{dx} = x^x (\ln x + 1)$

Example: Differentiate $y = (\sin x)^x$

$\ln y = x \ln (\sin x)$
$\frac{1}{y} \frac{dy}{dx} = (1) \ln (\sin x) + x \cdot \frac{\cos x}{\sin x} = \ln (\sin x) + x \cot x$
$\frac{dy}{dx} = (\sin x)^x (\ln (\sin x) + x \cot x)$

Tip: Always remember to substitute the original expression for $y$ in the final step. Forgetting this is a very common mistake!

Warning: Don't try to apply power rule or exponential rule directly to $u(x)^{v(x)}$ . They don't apply!

3. Parametric Differentiation: When $x$ and $y$ Depend on a Parameter

In parametric equations, both $x$ and $y$ are expressed as functions of a third variable, often denoted as $t$ . Think of $t$ as "time," and $x = f(t)$ and $y = g(t)$ describe the path of a particle in the $xy$ -plane as time evolves.

The Problem: How do we find $\frac{dy}{dx}$ when $y$ is not explicitly a function of $x$ ? We need a way to relate the rates of change of $x$ and $y$ with respect to the parameter $t>.

4. The Key Formula: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

If $x = f(t)$ and $y = g(t)$ , then:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$

Derivation (Intuition): Think of it as canceling out $dt$ in the fractions. Formally, it comes from the chain rule.

Example: Suppose $x = t^2$ and $y = 2t$ . Find $\frac{dy}{dx}$ .

$\frac{dx}{dt} = 2t$
$\frac{dy}{dt} = 2$
$\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}$

Interpretation: The slope of the curve defined by $x = t^2$ and $y = 2t$ at any point corresponding to parameter value $t$ is $\frac{1}{t}$ .

Example: Suppose $x = a \cos \theta$ and $y = a \sin \theta$ . Find $\frac{dy}{dx}$ .

$\frac{dx}{d\theta} = -a \sin \theta$
$\frac{dy}{d\theta} = a \cos \theta$
$\frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$

Tip: When finding higher-order derivatives (e.g., $\frac{d^2y}{dx^2}$ ), remember that $\frac{dy}{dx}$ is now a function of $t$ . You'll need to apply the chain rule again: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}$

Warning: Make sure you divide by $\frac{dx}{dt}$ , not $\frac{dt}{dx}$ . The order matters!

JEE Specific Tricks

Trick 1: Recognizing the Need for Logarithmic Differentiation Look for expressions with multiple terms multiplied/divided or variables in the exponent. This is a strong indicator that logarithmic differentiation will simplify the problem.

Trick 2: Chain Rule in Parametric Form JEE problems often combine parametric differentiation with other differentiation rules. Remember to apply the chain rule correctly when differentiating functions involving the parameter $t$ .

Trick 3: Implicit Differentiation Some problems require you to eliminate the parameter $t$ to express $y$ as a function of $x$ and then use implicit differentiation. Assess whether this elimination is possible before resorting to parametric differentiation.

You've now equipped yourself with the essential techniques of logarithmic and parametric differentiation. Practice these concepts with various problems to solidify your understanding. Good luck with your JEE preparation!