91Ó°ÊÓ

To find the first derivative, we will use the chain rule which states that \((f(g(x)))' = f'(g(x))g'(x)\). Let's identify our functions: Outer function: \(f(u) = e^u\) Inner function: \(g(x) = -2x^2\) Now we need to find their derivatives: \(f'(u) = e^u\) \(g'(x) = -4x\) Now using the chain rule: $$(f(g(x)))' = f'(g(x))g'(x) = e^{-2x^2}(-4x)$$ So, the first derivative is: $$\frac{dy}{dx} = -4xe^{-2x^2}$$

Again, we will use the chain rule to find the second derivative. This time, our function is \(\frac{dy}{dx} = -4xe^{-2x^2}\). Outer function: \(f(u) = -4xu\) Inner function: \(g(x) = e^{-2x^2}\) Find the derivatives of these functions: \(f'(u) = -4x\) (with respect to u) \(g'(x) = -4xe^{-2x^2}\) (we already found this in step 1) Now, using the chain rule: $$\frac{d^2y}{dx^2} = f'(g(x))g'(x) = (-4x)(-4xe^{-2x^2})$$ Now, we also need to take account of the derivative of the inner function with respect to x (chain rule): $$\frac{d^2y}{dx^2} = (-4x)(-4xe^{-2x^2}) + -4e^{-2x^2}(-4x)$$ Now simplify the expression: $$\frac{d^2y}{dx^2} = 16x^2e^{-2x^2} -16xe^{-2x^2}$$ So, the second derivative is: $$\frac{d^{2} y}{d x^{2}} = 16x^2e^{-2x^2} -16xe^{-2x^2}$$