Class CBSE Class 12 Mathematics Derivatives Q #1440
KNOWLEDGE BASED
UNDERSTAND
5 Marks 2025 AISSCE(Board Exam) LA
Find $\frac{dy}{dx}$ if $y^{x}+x^{y}+x^{x}=a^{b}$, where a and b are constants.
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Detailed Solution

Step 1: Define the functions

Let $u = y^x$, $v = x^y$, and $w = x^x$. Then the given equation can be written as $u + v + w = a^b$. Since $a$ and $b$ are constants, $a^b$ is also a constant. Therefore, the derivative of $a^b$ with respect to $x$ is 0.

Step 2: Differentiate the equation

Differentiating both sides of $u + v + w = a^b$ with respect to $x$, we get: $$\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0$$

Step 3: Find $\frac{du}{dx}$

Given $u = y^x$, take the natural logarithm of both sides: $$\ln u = \ln (y^x) = x \ln y$$ Differentiate both sides with respect to $x$: $$\frac{1}{u} \frac{du}{dx} = \ln y + x \cdot \frac{1}{y} \frac{dy}{dx}$$ $$\frac{du}{dx} = u \left( \ln y + \frac{x}{y} \frac{dy}{dx} \right) = y^x \left( \ln y + \frac{x}{y} \frac{dy}{dx} \right)$$

Step 4: Find $\frac{dv}{dx}$

Given $v = x^y$, take the natural logarithm of both sides: $$\ln v = \ln (x^y) = y \ln x$$ Differentiate both sides with respect to $x$: $$\frac{1}{v} \frac{dv}{dx} = \frac{dy}{dx} \ln x + y \cdot \frac{1}{x}$$ $$\frac{dv}{dx} = v \left( \ln x \frac{dy}{dx} + \frac{y}{x} \right) = x^y \left( \ln x \frac{dy}{dx} + \frac{y}{x} \right)$$

Step 5: Find $\frac{dw}{dx}$

Given $w = x^x$, take the natural logarithm of both sides: $$\ln w = \ln (x^x) = x \ln x$$ Differentiate both sides with respect to $x$: $$\frac{1}{w} \frac{dw}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$$ $$\frac{dw}{dx} = w (\ln x + 1) = x^x (\ln x + 1)$$

Step 6: Substitute the derivatives into the equation

Substituting $\frac{du}{dx}$, $\frac{dv}{dx}$, and $\frac{dw}{dx}$ into $\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0$, we get: $$y^x \left( \ln y + \frac{x}{y} \frac{dy}{dx} \right) + x^y \left( \ln x \frac{dy}{dx} + \frac{y}{x} \right) + x^x (\ln x + 1) = 0$$

Step 7: Solve for $\frac{dy}{dx}$

Rearrange the equation to isolate $\frac{dy}{dx}$: $$y^x \ln y + y^x \frac{x}{y} \frac{dy}{dx} + x^y \ln x \frac{dy}{dx} + x^y \frac{y}{x} + x^x (\ln x + 1) = 0$$ $$\frac{dy}{dx} \left( y^x \frac{x}{y} + x^y \ln x \right) = -y^x \ln y - x^y \frac{y}{x} - x^x (\ln x + 1)$$ $$\frac{dy}{dx} = \frac{-y^x \ln y - x^y \frac{y}{x} - x^x (\ln x + 1)}{y^x \frac{x}{y} + x^y \ln x}$$ $$\frac{dy}{dx} = -\frac{y^x \ln y + yx^{y-1} + x^x (\ln x + 1)}{xy^{x-1} + x^y \ln x}$$

Final Answer: $$-\frac{y^x \ln y + yx^{y-1} + x^x (\ln x + 1)}{xy^{x-1} + x^y \ln x}$$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concepts of implicit differentiation and logarithmic differentiation to solve the problem. They must also understand how to apply the chain rule and product rule correctly.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply specific procedures like logarithmic differentiation, chain rule, and product rule to find the derivative. It's about knowing 'how' to differentiate rather than just recalling facts or understanding concepts in isolation.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of differentiation techniques covered in the textbook, specifically implicit and logarithmic differentiation.

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