1. **Problem statement:** Given the curve equation $$x = \frac{1}{5} \left( e^{y(2x-3)} + 4 \right)$$, find the value of $$\frac{dy}{dx}$$ when $$x=1$$. 2. **Step 1: Differentiate both sides with respect to $$x$$.** We have $$x$$ on the left and a function of $$x$$ and $$y$$ on the right. Use implicit differentiation. $$\frac{d}{dx} \left( x \right) = \frac{d}{dx} \left( \frac{1}{5} \left( e^{y(2x-3)} + 4 \right) \right)$$ 3. **Step 2: Differentiate the left side:** $$\frac{d}{dx} (x) = 1$$ 4. **Step 3: Differentiate the right side:** $$\frac{d}{dx} \left( \frac{1}{5} e^{y(2x-3)} + \frac{4}{5} \right) = \frac{1}{5} \frac{d}{dx} \left( e^{y(2x-3)} \right) + 0$$ 5. **Step 4: Differentiate $$e^{y(2x-3)}$$ using chain rule:** Let $$u = y(2x-3)$$, then $$\frac{d}{dx} e^u = e^u \frac{du}{dx}$$ 6. **Step 5: Differentiate $$u = y(2x-3)$$ using product rule:** $$\frac{du}{dx} = y \frac{d}{dx} (2x-3) + (2x-3) \frac{dy}{dx} = y \cdot 2 + (2x-3) \frac{dy}{dx} = 2y + (2x-3) \frac{dy}{dx}$$ 7. **Step 6: Substitute back:** $$\frac{d}{dx} e^{y(2x-3)} = e^{y(2x-3)} \left( 2y + (2x-3) \frac{dy}{dx} \right)$$ 8. **Step 7: Write the full differentiated equation:** $$1 = \frac{1}{5} e^{y(2x-3)} \left( 2y + (2x-3) \frac{dy}{dx} \right)$$ 9. **Step 8: Solve for $$\frac{dy}{dx}$$:** Multiply both sides by 5: $$5 = e^{y(2x-3)} \left( 2y + (2x-3) \frac{dy}{dx} \right)$$ Isolate terms with $$\frac{dy}{dx}$$: $$5 = 2y e^{y(2x-3)} + (2x-3) e^{y(2x-3)} \frac{dy}{dx}$$ Subtract $$2y e^{y(2x-3)}$$: $$5 - 2y e^{y(2x-3)} = (2x-3) e^{y(2x-3)} \frac{dy}{dx}$$ Divide both sides by $$(2x-3) e^{y(2x-3)}$$: $$\frac{5 - 2y e^{y(2x-3)}}{(2x-3) e^{y(2x-3)}} = \frac{dy}{dx}$$ Use cancellation notation: $$\frac{\cancel{5} - 2y \cancel{e^{y(2x-3)}}}{(2x-3) \cancel{e^{y(2x-3)}}} = \frac{dy}{dx}$$ 10. **Step 9: Find $$y$$ when $$x=1$$ using original equation:** $$1 = \frac{1}{5} \left( e^{y(2(1)-3)} + 4 \right) = \frac{1}{5} \left( e^{y(-1)} + 4 \right)$$ Multiply both sides by 5: $$5 = e^{-y} + 4$$ Subtract 4: $$1 = e^{-y}$$ Take natural log: $$\ln(1) = -y \Rightarrow 0 = -y \Rightarrow y=0$$ 11. **Step 10: Substitute $$x=1$$ and $$y=0$$ into $$\frac{dy}{dx}$$ formula:** $$\frac{dy}{dx} = \frac{5 - 2 \cdot 0 \cdot e^{0}}{(2 \cdot 1 - 3) e^{0}} = \frac{5 - 0}{(2 - 3) \cdot 1} = \frac{5}{-1} = -5$$ **Final answer:** $$\boxed{\frac{dy}{dx} = -5 \text{ when } x=1}$$