1. **State the problem:** We need to find the second derivative with respect to $x$ of the function $$f(x) = e^{7x-4}$$. 2. **Recall the formula:** The derivative of an exponential function $e^{u(x)}$ is given by $$\frac{d}{dx} e^{u(x)} = u'(x) e^{u(x)}$$ where $u'(x)$ is the derivative of the exponent. 3. **First derivative:** Let $u(x) = 7x - 4$, so $$u'(x) = 7$$. Therefore, the first derivative is: $$\frac{d}{dx} e^{7x-4} = 7 e^{7x-4}$$ 4. **Second derivative:** Now differentiate the first derivative again: $$\frac{d^2}{dx^2} e^{7x-4} = \frac{d}{dx} \left(7 e^{7x-4}\right) = 7 \frac{d}{dx} e^{7x-4}$$ Using the derivative rule again: $$= 7 \times 7 e^{7x-4} = 49 e^{7x-4}$$ 5. **Final answer:** $$\boxed{\frac{d^2}{dx^2} e^{7x-4} = 49 e^{7x-4}}$$