If x=e^theta(sin theta-cos theta) and y=e^the | vedclass

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(A) Given $x=e^	heta(\sin 	heta-\cos 	heta)$ and $y=e^	heta(\sin 	heta+\cos 	heta)$.Differentiating $x$ with respect to $	heta$ using the produ

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(A) Given $x=e^	heta(\sin 	heta-\cos 	heta)$ and $y=e^	heta(\sin 	heta+\cos 	heta)$.Differentiating $x$ with respect to $	heta$ using the product rule:$\frac{dx}{d	heta} = e^	heta(\cos 	heta + \sin 	heta) + e^	heta(\sin 	heta - \cos 	heta) = e^	heta(\cos 	heta + \sin 	heta + \sin 	heta - \cos 	heta) = 2e^	heta \sin 	heta$.Differentiating $y$ with respect to $	heta$ using the product rule:$\frac{dy}{d	heta} = e^	heta(\cos 	heta - \sin 	heta) + e^	heta(\sin 	heta + \cos 	heta) = e^	heta(\cos 	heta - \sin 	heta + \sin 	heta + \cos 	heta) = 2e^	heta \cos 	heta$.Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{2e^	heta \cos 	heta}{2e^	heta \sin 	heta} = \cot 	heta$.Evaluating at $	heta = \frac{\pi}{4}$:$\left. \frac{dy}{dx} ight|_{	heta=\frac{\pi}{4}} = \cot \left( \frac{\pi}{4} ight) = 1$.

If $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$,then $\frac{dy}{dx}$ at $\theta=\frac{\pi}{4}$ is:

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