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I'm having a bit of trouble on this one problem

Find the slope of the tangent line to the curve

$$\cos(3x−2y)−xe^{−x}=−\dfrac{5π}2e^{−5π/2}.$$

At the point $\left(\tfrac{5\pi}2,4\pi\right)$.

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To find the slope at that point, you need $\frac{dy}{dx}$. So differentiate both sides of your equation with respect to $x$.

$\frac{d}{dx}\big(\cos(3x-2y)-xe^{-x}\big) = \frac{d}{dx}\big(-\frac{5\pi}{2}e^{-5\pi/2}\big)$

$\frac{d}{dx}(\cos(3x-2y))-\frac{d}{dx}(xe^{-x}) = \frac{d}{dx}\big(-\frac{5\pi}{2}e^{-5\pi/2}\big)$

$-\sin(3x-2y)\frac{d}{dx}(3x-2y)-\frac{d}{dx}(xe^{-x})$=0

$-\sin(3x-2y)(3-2\frac{dx}{dy})-e^{-x}+xe^{-x}=0$

And so, solving for $\frac{dy}{dx}$, we have that:

$\frac{dy}{dx}=\frac{e^{-x}+xe^{-x}}{2\sin(3x-2y)}+\frac{3}{2}$

Now plug in your $x=\frac{5\pi}{2}$ and $y=4\pi$ to get $\frac{dy}{dx}$

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