Implicit Differentiation for (cos x)^y = (sin y)^x to Find dy/dx

Math Problem Statement

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If (cos x)^y = (sin y)^x, then find dy/dx.

To solve for $\frac{dy}{dx}$ given the equation $(\cos x)^y = (\sin y)^x$ , we will proceed with logarithmic differentiation.

Given: $(\cos x)^y = (\sin y)^x$ take the natural logarithm on both sides: $\ln\left((\cos x)^y\right) = \ln\left((\sin y)^x\right)$

Using the power rule of logarithms, $\ln(a^b) = b \ln(a)$ , we get: $y \ln(\cos x) = x \ln(\sin y)$

Now we apply implicit differentiation with respect to $x$ .

For the left side:

= -y \tan x + \ln(\cos x) \cdot \frac{dy}{dx}$$ For the right side: $$\frac{d}{dx}(x \ln(\sin y)) = \ln(\sin y) + x \cdot \frac{\cos y}{\sin y} \cdot \frac{dy}{dx} = \ln(\sin y) + x \cot y \cdot \frac{dy}{dx}$$ ### Step 3: Set up the equation Equating both differentiated sides: $$-y \tan x + \ln(\cos x) \cdot \frac{dy}{dx} = \ln(\sin y) + x \cot y \cdot \frac{dy}{dx}$$ ### Step 4: Solve for $$\frac{dy}{dx}$$ Rearrange terms to isolate $$\frac{dy}{dx}$$: $$\frac{dy}{dx} \left( \ln(\cos x) - x \cot y \right) = \ln(\sin y) + y \tan x$$ Thus, $$\frac{dy}{dx} = \frac{\ln(\sin y) + y \tan x}{\ln(\cos x) - x \cot y}$$ ### Final Answer: $$\frac{dy}{dx} = \frac{\ln(\sin y) + y \tan x}{\ln(\cos x) - x \cot y}$$ Let me know if you have any questions or would like further clarification. --- ### Related Questions 1. How would the approach change if $$(\cos x)^y$$ were replaced by $$(\sin x)^y$$? 2. What other methods could we use to differentiate equations of this form? 3. How does the chain rule apply in this type of problem? 4. What are some common pitfalls when performing implicit differentiation? 5. How can logarithmic differentiation simplify problems involving powers and roots? ### Tip Always apply logarithmic differentiation when you have variables in both the base and the exponent, as it simplifies the differentiation process significantly.

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Calculus
Implicit Differentiation
Logarithmic Differentiation

Power Rule of Logarithms: ln(a^b) = b ln(a)
Implicit Differentiation

Chain Rule

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