Class CBSE Class 12 Mathematics Derivatives Q #1340
KNOWLEDGE BASED
UNDERSTAND
3 Marks 2024 AISSCE(Board Exam) SA
Find $\frac{dy}{dx}$ , if $(cos~x)^{y}=(cos~y)^{x}.$
Prev Next

AI Tutor Explanation

Powered by Gemini

Detailed Solution

Step 1: Apply Logarithm on both sides

Given the equation $(cos~x)^{y}=(cos~y)^{x}$, we take the natural logarithm of both sides to simplify the exponents. $ln((cos~x)^{y}) = ln((cos~y)^{x})$ Using the power rule of logarithms, we get: $y~ln(cos~x) = x~ln(cos~y)$

Step 2: Differentiate both sides with respect to x

Now, we differentiate both sides of the equation $y~ln(cos~x) = x~ln(cos~y)$ with respect to $x$ using the product rule and chain rule. $\frac{d}{dx}[y~ln(cos~x)] = \frac{d}{dx}[x~ln(cos~y)]$ Applying the product rule on both sides: $\frac{dy}{dx}ln(cos~x) + y\frac{d}{dx}ln(cos~x) = ln(cos~y) + x\frac{d}{dx}ln(cos~y)$

Step 3: Apply Chain Rule

Now, we apply the chain rule to differentiate $ln(cos~x)$ and $ln(cos~y)$ with respect to $x$. $\frac{d}{dx}ln(cos~x) = \frac{1}{cos~x} * (-sin~x) = -tan~x$ $\frac{d}{dx}ln(cos~y) = \frac{1}{cos~y} * (-sin~y) * \frac{dy}{dx} = -tan~y * \frac{dy}{dx}$ Substituting these derivatives back into the equation: $\frac{dy}{dx}ln(cos~x) + y(-tan~x) = ln(cos~y) + x(-tan~y)\frac{dy}{dx}$ $\frac{dy}{dx}ln(cos~x) - y~tan~x = ln(cos~y) - x~tan~y~\frac{dy}{dx}$

Step 4: Rearrange and solve for dy/dx

Now, we rearrange the equation to isolate $\frac{dy}{dx}$ terms on one side: $\frac{dy}{dx}ln(cos~x) + x~tan~y~\frac{dy}{dx} = ln(cos~y) + y~tan~x$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}[ln(cos~x) + x~tan~y] = ln(cos~y) + y~tan~x$ Finally, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{ln(cos~y) + y~tan~x}{ln(cos~x) + x~tan~y}$

Final Answer: $\frac{dy}{dx} = \frac{ln(cos~y) + y~tan~x}{ln(cos~x) + x~tan~y}$

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concepts of logarithmic differentiation, product rule, and chain rule to solve the problem. They must apply these rules in a specific order to arrive at the correct derivative.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (logarithmic differentiation, product rule, chain rule) to find the derivative. It's about knowing how to do something rather than just knowing facts or concepts.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge and application of differentiation techniques, which are core concepts in the syllabus. The question is a standard application of logarithmic differentiation.

More from this Chapter

VSA
If $-2x^{2}-5xy+y^{3}=76$, then find $\frac{dy}{dx}$.
MCQ_SINGLE
The derivative of \(2^{x}\) w.r.t. \(3^{x}\) is:
SA
Differentiate $y=\sqrt{\log\left\{\sin\left(\frac{x^{3}}{3}-1\right)\right\}}$ with respect to x.
VSA
If $f(x)=|tan~2x|$, then find the value of $f^{\prime}(x)$ at $x=\frac{\pi}{3}$
SA
If $x=e^{cos~3t}$ and $y=e^{sin~3t}$ , prove that $\frac{dy}{dx}=-\frac{y~log~x}{x~log~y}$
View All Questions