91Ó°ÊÓ

To find the first derivative, we'll apply the chain rule. The chain rule states that if \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). Here, our function \(y = e^{-2 x^{2}}\), which can be written as \(y = f(g(x))\) where \(f(u) = e^{u}\) and \(g(x)=-2x^{2}\). First, let's find the derivatives of \(f(u)\) and \(g(x)\): $$f'(u) = \frac{d}{du}(e^{u}) = e^{u}$$ $$g'(x) = \frac{d}{dx}(-2x^{2}) = -4x$$ Now, let's apply the chain rule to find the first derivative of \(y\): $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = e^{-2x^{2}} \cdot (-4x)$$ So, the first derivative is given by: $$\frac{dy}{dx} = -4xe^{-2x^{2}}$$

Now, we need to find the second derivative, which means finding the derivative of the first derivative. So, we need to differentiate \(-4xe^{-2x^{2}}\) with respect to \(x\). To do this, we shall use the product rule. The product rule states that if \(h(x) = v(x)w(x)\), then \(\frac{d}{dx}(h(x)) = v'(x)w(x) + v(x)w'(x)\). Here, \(v(x) = -4x\) and \(w(x) = e^{-2 x^{2}}\). First, let's find the derivatives of \(v(x)\) and \(w(x)\): $$v'(x) = \frac{d}{dx}(-4x) = -4$$ $$w'(x) = \frac{d}{dx}(e^{-2 x^{2}}) = e^{-2 x^{2}} \cdot (-4x)$$ (We already found this in Step 1 as the first derivative of the function) Now, let's apply the product rule to find the second derivative: $$\frac{d^{2} y}{d x^{2}} = v'(x)w(x) + v(x)w'(x) = (-4)e^{-2x^{2}} + (-4x)(-4xe^{-2x^{2}})$$ Finally, simplify the expression: $$\frac{d^{2} y}{d x^{2}} = -4e^{-2x^{2}} + 16x^{2}e^{-2x^{2}}$$ So the second derivative of the given function is: $$\frac{d^{2} y}{d x^{2}} = -4e^{-2x^{2}}(1 - 4x^{2})$$