How to differentiate $y=e^ {\frac{-1}{2}x^2}\cdot\sqrt{1+2x^2}$?

Question

We need to find the maximum point of the curve, M. I know that we must find the $dy/dx$ of the equation of the curve and set it to $0$. However, I'm having trouble differentiating the equation. I'm aware of the product/quotient rule, but didn't get the correct answer despite trying several times.

Answer 1

Use the product rule, $$\frac{d}{dx}f(x)g(x) = f'(x)g(x)+f(x)g'(x)$$

Here, $$f(x) = e^{-\frac{1}{2}x^2}$$ $$g(x) = \sqrt{1+2x^2}$$

To find $f'(x)$ and $g'(x)$ use the chain rule,

$$f'(x) = e^{-\frac{1}{2}x^2} \cdot (-x)$$ $$g'(x) = \frac{4x}{2\sqrt{1+2x^2}}$$

Answer 2

HINT: $$(e^{-1/2x^2})'=e^{-1/2x^2} \cdot (-x)$$ and $$(\sqrt{1+2x^2})'=\frac{1}{2}(1+2x^2)^{-1/2}\cdot 4x$$ and the first derivative is given by $$-{\frac {x{{\rm e}^{-1/2\,{x}^{2}}} \left( 2\,{x}^{2}-1 \right) }{ \sqrt {2\,{x}^{2}+1}}} $$ (simplified)