Question

If \( x = e^{\cos 3t} \) and \( y = e^{\sin 3t} \), prove that \( \frac{dy}{dx} = -\frac{y \log x}{x \log y} \).

Hint:

When \( x \) and \( y \) are exponential functions, logarithmic identities are often helpful in simplifying derivatives.

Answer 1

Step 1: Differentiate \( x = e^{\cos 3t} \).
The given equation is \( x = e^{\cos 3t} \). Taking the derivative with respect to \( t \): \[ \frac{dx}{dt} = e^{\cos 3t} \cdot \frac{d}{dt}(\cos 3t). \] Using the chain rule: \[ \frac{dx}{dt} = e^{\cos 3t} \cdot (-\sin 3t) \cdot 3. \] Simplify: \[ \frac{dx}{dt} = -3x \sin 3t. \] Step 2: Differentiate \( y = e^{\sin 3t} \).
The given equation is \( y = e^{\sin 3t} \). Taking the derivative with respect to \( t \): \[ \frac{dy}{dt} = e^{\sin 3t} \cdot \frac{d}{dt}(\sin 3t). \] Using the chain rule: \[ \frac{dy}{dt} = e^{\sin 3t} \cdot (\cos 3t) \cdot 3. \] Simplify: \[ \frac{dy}{dt} = 3y \cos 3t. \] Step 3: Compute \( \frac{dy}{dx} \).
The chain rule gives: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \] Substitute \( \frac{dy}{dt} = 3y \cos 3t \) and \( \frac{dx}{dt} = -3x \sin 3t \): \[ \frac{dy}{dx} = \frac{3y \cos 3t}{-3x \sin 3t}. \] Simplify: \[ \frac{dy}{dx} = -\frac{y \cos 3t}{x \sin 3t}. \] Step 4: Relate to logarithms.
From the given equations: \[ \cos 3t = \log x \quad \text{and} \quad \sin 3t = \log y. \] Substitute \( \cos 3t = \log x \) and \( \sin 3t = \log y \) into \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{y \log x}{x \log y}. \] Conclusion:
The result is proven: \[ \boxed{\frac{dy}{dx} = -\frac{y \log x}{x \log y}}. \]