Gradient Curve 11371B

1. **State the problem:** We need to find the gradient (derivative) of the curve given by $$y = (e^{2x} - e^{-2x})^2$$ at the point where $$x = \ln 3$$. 2. **Formula and rules:** The gradient of a curve at a point is given by the derivative $$\frac{dy}{dx}$$ evaluated at that point. 3. **Differentiate the function:** Use the chain rule. Let $$u = e^{2x} - e^{-2x}$$, so $$y = u^2$$. Then, $$\frac{dy}{dx} = 2u \cdot \frac{du}{dx}$$. Calculate $$\frac{du}{dx}$$: $$\frac{du}{dx} = \frac{d}{dx}(e^{2x} - e^{-2x}) = 2e^{2x} + 2e^{-2x}$$. 4. **Substitute back:** $$\frac{dy}{dx} = 2(e^{2x} - e^{-2x})(2e^{2x} + 2e^{-2x}) = 4(e^{2x} - e^{-2x})(e^{2x} + e^{-2x})$$. 5. **Simplify the expression:** Use the identity: $$(a - b)(a + b) = a^2 - b^2$$ So, $$\frac{dy}{dx} = 4(e^{4x} - e^{-4x})$$. 6. **Evaluate at $$x = \ln 3$$:** Calculate: $$e^{4 \ln 3} = (e^{\ln 3})^4 = 3^4 = 81$$ $$e^{-4 \ln 3} = (e^{\ln 3})^{-4} = 3^{-4} = \frac{1}{81}$$ Therefore, $$\frac{dy}{dx} \bigg|_{x=\ln 3} = 4(81 - \frac{1}{81}) = 4\left(\frac{6561}{81} - \frac{1}{81}\right) = 4 \cdot \frac{6560}{81} = \frac{26240}{81}$$. 7. **Final answer:** The gradient of the curve at $$x = \ln 3$$ is $$\boxed{\frac{26240}{81}}$$.