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Continuity and Differentiability - Logarithmic differentiation

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Logarithmic differentiation is a technique used to differentiate functions that are products of many terms, quotients of complex expressions, or functions where the variable is in the exponent, such as [f(x)]g(x)[f(x)]^{g(x)}. Visually, this method 'unpacks' complex nested powers into simpler product forms.

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The core idea involves taking the natural logarithm (base ee) on both sides of an equation y=f(x)y = f(x) before differentiating. This utilizes the properties of logarithms to transform powers into products and products into sums, simplifying the differentiation process.

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When applying this to a power function y=uvy = u^v, taking the log transforms the expression to log⁡y=vlog⁡u\log y = v \log u. Visually, the exponent vv 'drops down' to become a multiplier, allowing the use of the Product Rule instead of complex power rules.

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The derivative of log⁡y\log y with respect to xx is always 1y⋅dydx\frac{1}{y} \cdot \frac{dy}{dx} due to the Chain Rule. This step is crucial because it requires multiplying the entire RHS by yy at the end to isolate dydx\frac{dy}{dx}.

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For functions involving sums or differences like y=u(x)+v(x)y = u(x) + v(x) where uu and vv are power functions, you cannot take the log of the whole side immediately (since log⁡(u+v)≠log⁡u+log⁡v\log(u+v) \neq \log u + \log v). Instead, differentiate uu and vv separately using logarithmic differentiation and then add their derivatives.

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The domain of the function must be considered; since log⁡x\log x is only defined for x>0x > 0, we typically assume the functions involved are positive or use absolute values ∣f(x)∣|f(x)| to ensure the logarithm is well-defined over the required interval.

📐Formulae

Product Rule for Logs: log⁡(m⋅n)=log⁡m+log⁡n\log(m \cdot n) = \log m + \log n

Quotient Rule for Logs: log⁡(mn)=log⁡m−log⁡n\log(\frac{m}{n}) = \log m - \log n

Power Rule for Logs: log⁡(mn)=nlog⁡m\log(m^n) = n \log m

Derivative of log: ddx(log⁡x)=1x\frac{d}{dx}(\log x) = \frac{1}{x}

Chain Rule for log yy: ddx(log⁡y)=1ydydx\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}

General Result: If y=[f(x)]g(x)y = [f(x)]^{g(x)}, then dydx=[f(x)]g(x)[g′(x)log⁡f(x)+g(x)f′(x)f(x)]\frac{dy}{dx} = [f(x)]^{g(x)} [g'(x) \log f(x) + g(x) \frac{f'(x)}{f(x)}]

💡Examples

Problem 1:

Differentiate y=xsin⁡xy = x^{\sin x} with respect to xx.

Solution:

Step 1: Take natural log on both sides: log⁡y=log⁡(xsin⁡x)\log y = \log(x^{\sin x}) Step 2: Use the power property of logarithms: log⁡y=sin⁡x⋅log⁡x\log y = \sin x \cdot \log x Step 3: Differentiate both sides with respect to xx: 1ydydx=ddx(sin⁡x⋅log⁡x)\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x \cdot \log x) Step 4: Apply the Product Rule on the RHS: 1ydydx=sin⁡x⋅ddx(log⁡x)+log⁡x⋅ddx(sin⁡x)\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(\sin x) 1ydydx=sin⁡x⋅1x+log⁡x⋅cos⁡x\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{1}{x} + \log x \cdot \cos x Step 5: Solve for dydx\frac{dy}{dx} by multiplying by yy: dydx=y[sin⁡xx+cos⁡xlog⁡x]\frac{dy}{dx} = y [\frac{\sin x}{x} + \cos x \log x] Step 6: Substitute the original value of yy: dydx=xsin⁡x[sin⁡xx+cos⁡xlog⁡x]\frac{dy}{dx} = x^{\sin x} [\frac{\sin x}{x} + \cos x \log x].

Explanation:

This approach uses logarithmic differentiation to handle the variable in the exponent. By converting the power to a product, we can apply standard differentiation rules like the Product Rule.

Problem 2:

Find dydx\frac{dy}{dx} if y=xx+ax+xa+aay = x^x + a^x + x^a + a^a.

Solution:

Let u=xxu = x^x. Then log⁡u=xlog⁡x\log u = x \log x. Differentiating gives 1ududx=1⋅log⁡x+x⋅1x=log⁡x+1\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1. So, dudx=xx(1+log⁡x)\frac{du}{dx} = x^x(1 + \log x). Now, for the other terms: ddx(ax)=axlog⁡a\frac{d}{dx}(a^x) = a^x \log a ddx(xa)=axa−1\frac{d}{dx}(x^a) = a x^{a-1} ddx(aa)=0\frac{d}{dx}(a^a) = 0 (since it is a constant) Combining all parts: dydx=xx(1+log⁡x)+axlog⁡a+axa−1\frac{dy}{dx} = x^x(1 + \log x) + a^x \log a + a x^{a-1}.

Explanation:

This example demonstrates that logarithmic differentiation is applied individually to complex power terms within a sum. It also highlights the difference between variable-to-variable, constant-to-variable, and variable-to-constant differentiation.