Continuity and Differentiability Ex 5.5 Class 12 NCERT

Core Concept: What is Logarithmic Differentiation?

Logarithmic differentiation is a technique used to differentiate functions that are either products of several factors, quotients of complex terms, or of the form $y = [f(x)]^{g(x)}$, where both the base and the exponent are functions of $x$. Standard rules like the power rule ($d/dx(x^n) = n x^{n-1}$) only apply when the exponent is a constant. Similarly, exponential rules ($d/dx(a^x) = a^x \ln a$) only apply when the base is a constant.

When both base and exponent are variables, we must use logarithms to pull the exponent down. By taking the natural logarithm (base $e$, denoted as $\ln$ or $\log$) on both sides, we convert the exponentiation into multiplication: $\ln(y) = g(x) \ln[f(x)]$. We then use implicit differentiation on the left side to get $\frac{1}{y} \frac{dy}{dx}$, and apply the product rule on the right side. This technique converts complex, multi-factor algebraic fractions into simple additions and subtractions using the core logarithmic identities:

1. $\log(ab) = \log a + \log b$
2. $\log(a/b) = \log a - \log b$
3. $\log(a^b) = b \log a$

The Step-by-Step Logarithmic Differentiation Method

1. Equate and Take Natural Logarithm — Set your given function equal to $y$. Take the natural logarithm ($\log$ to the base $e$) on both sides of the equation.
2. Apply Logarithmic Properties — Simplify the expression on the right-hand side using the laws of logarithms. Convert powers to multipliers, products to sums, and quotients to differences.
3. Differentiate Implicitly — Differentiate both sides with respect to $x$. Remember that the derivative of $\log y$ with respect to $x$ is $\frac{1}{y} \frac{dy}{dx}$ (using the Chain Rule). Differentiate the right side using standard rules like the product rule or chain rule.
4. Isolate dy/dx and Substitute y — Multiply both sides by $y$ to solve for $\frac{dy}{dx}$. Finally, replace $y$ with the original function in terms of $x$ so that your final answer is entirely in terms of $x$.

Fully Worked-Out NCERT Examples

• Example 1: Differentiate $y = x^x$ with respect to $x$. Step 1: Take natural log on both sides: $\log y = \log(x^x) \Rightarrow \log y = x \log x$. Step 2: Differentiate both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[x \log x]$. Step 3: Apply the product rule on the right-hand side: $\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) = x \left(\frac{1}{x}\right) + \log x (1) = 1 + \log x$. Step 4: Multiply by $y$: $\frac{dy}{dx} = y(1 + \log x)$. Step 5: Substitute $y = x^x$ back: $\frac{dy}{dx} = x^x(1 + \log x)$.
• Example 2: Differentiate $y = x^{\sin x}$ with respect to $x$. Step 1: Take natural log on both sides: $\log y = \log(x^{\sin x}) \Rightarrow \log y = \sin x \log x$. Step 2: Differentiate both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[\sin x \log x]$. Step 3: Apply product rule on the right-hand side: $\frac{1}{y} \frac{dy}{dx} = \sin x \left(\frac{1}{x}\right) + \log x (\cos x)$. Step 4: Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = y \left[ \frac{\sin x}{x} + \cos x \log x \right]$. Step 5: Substitute $y = x^{\sin x}$ back: $\frac{dy}{dx} = x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \log x \right]$.

CBSE Board Exam Trap: The 'Sum of Functions' Error

One of the most common mistakes students make in Ex 5.5 is when differentiating functions of the form $y = u(x) + v(x)$, such as $y = x^{\sin x} + (\sin x)^x$.

The Trap: Students write $\log y = \log(u + v) = \log u + \log v$. This is algebraically incorrect! There is no property of logarithms that allows you to expand $\log(u + v)$. Doing this will result in zero marks for the question.

The Correct Approach: You must define individual variables for each part: $y = u + v$, where $u = x^{\sin x}$ and $v = (\sin x)^x$. Differentiate them separately to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ using logarithmic differentiation, and then add them together: $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.

Practice Questions with Solutions

• Q: Differentiate $(\log x)^{\cos x}$ with respect to $x$. A: Step 1: Let $y = (\log x)^{\cos x}$. Step 2: Taking natural logarithm on both sides: $\log y = \log[(\log x)^{\cos x}] \Rightarrow \log y = \cos x \cdot \log(\log x)$. Step 3: Differentiating both sides with respect to $x$ using the Chain Rule and Product Rule: $\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \frac{d}{dx}[\log(\log x)] + \log(\log x) \cdot \frac{d}{dx}(\cos x)$ $\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \left(\frac{1}{\log x} \cdot \frac{1}{x}\right) + \log(\log x) \cdot (-\sin x)$ $\frac{1}{y} \frac{dy}{dx} = \frac{\cos x}{x \log x} - \sin x \log(\log x)$. Step 4: Multiply by $y$ and substitute $y = (\log x)^{\cos x}$: $\frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right]$. Final answer: $\frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right]$
• Q: If $x^y = e^{x-y}$, prove that $\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}$. A: Step 1: Given equation: $x^y = e^{x-y}$. Step 2: Taking natural logarithm on both sides: $\log(x^y) = \log(e^{x-y}) \Rightarrow y \log x = (x-y) \log e$. Since $\log e = 1$, we get: $y \log x = x - y$. Step 3: Express $y$ explicitly in terms of $x$ to make differentiation simple: $y \log x + y = x \Rightarrow y(1 + \log x) = x \Rightarrow y = \frac{x}{1 + \log x}$. Step 4: Differentiate both sides with respect to $x$ using the Quotient Rule: $\frac{dy}{dx} = \frac{(1 + \log x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(1 + \log x)}{(1 + \log x)^2}$ $\frac{dy}{dx} = \frac{(1 + \log x)(1) - x \left(0 + \frac{1}{x}\right)}{(1 + \log x)^2}$ $\frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2} = \frac{\log x}{(1 + \log x)^2}$. Hence Proved. Final answer: Proven using the quotient rule after isolating $y$.
• Q: Find $\frac{dy}{dx}$ if $x^y + y^x = 1$. A: Step 1: Let $u = x^y$ and $v = y^x$. The given equation is $u + v = 1$. Differentiating both sides with respect to $x$ gives: $\frac{du}{dx} + \frac{dv}{dx} = 0$. Step 2: Find $\frac{du}{dx}$ using $u = x^y$: $\log u = y \log x \Rightarrow \frac{1}{u} \frac{du}{dx} = y \left(\frac{1}{x}\right) + \log x \frac{dy}{dx}$ $\frac{du}{dx} = x^y \left( \frac{y}{x} + \log x \frac{dy}{dx} \right) = y x^{y-1} + x^y \log x \frac{dy}{dx}$. Step 3: Find $\frac{dv}{dx}$ using $v = y^x$: $\log v = x \log y \Rightarrow \frac{1}{v} \frac{dv}{dx} = x \left(\frac{1}{y} \frac{dy}{dx}\right) + \log y (1)$ $\frac{dv}{dx} = y^x \left( \frac{x}{y} \frac{dy}{dx} + \log y \right) = x y^{x-1} \frac{dy}{dx} + y^x \log y$. Step 4: Substitute $\frac{du}{dx}$ and $\frac{dv}{dx}$ back into the equation: $\left( y x^{y-1} + x^y \log x \frac{dy}{dx} \right) + \left( x y^{x-1} \frac{dy}{dx} + y^x \log y \right) = 0$ $\frac{dy}{dx} [ x^y \log x + x y^{x-1} ] = - (y x^{y-1} + y^x \log y)$ $\frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}$. Final answer: $\frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}}$
• Q: Differentiate $y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$ with respect to $x$. A: Step 1: Rewrite the equation as $y = \left[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right]^{1/2}$. Step 2: Take natural logarithm on both sides: $\log y = \frac{1}{2} \log \left[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right]$ $\log y = \frac{1}{2} \left[ \log(x-1) + \log(x-2) - \log(x-3) - \log(x-4) - \log(x-5) \right]$. Step 3: Differentiate both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$. Step 4: Multiply by $y$ and substitute back: $\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$. Final answer: $\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$

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