Question

Find \(\frac{d^{2} y}{d x^{2}}\) for the following functions. $$y=e^{-2 x^{2}}$$

Short answer

Question: Find the second derivative of the function \(y = e^{-2 x^{2}}\). Answer: The second derivative of the function \(y = e^{-2 x^{2}}\) is \(\frac{d^{2} y}{d x^{2}} = -4e^{-2x^{2}}(1 - 4x^{2})\).

Step-by-step solution

Find the first derivative of the function

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: $$\frac{dy}{dx} = -4xe^{-2x^{2}}$$

Find the second derivative of the function

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: $$\frac{d^{2} y}{d x^{2}} = -4e^{-2x^{2}}(1 - 4x^{2})$$

Key concepts

Chain Rule

The chain rule is an essential tool in calculus when dealing with composite functions. A composite function is when one function is nested inside another, like our given function, where we have an exponential function raised to a negative quadratic power.

To apply the chain rule, identify the outer function \(f(u)\) and the inner function \(g(x)\). For instance, in this exercise:

• The outer function is \(f(u) = e^{u}\)
• The inner function is \(g(x) = -2x^{2}\)

Then differentiate each separately, remembering:

• The derivative of the outer function \(f'(u)=e^u\).
• The derivative of the inner function \(g'(x) = -4x\).

Join these using the chain rule formula: \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). This gives: \(-4xe^{-2x^{2}}\), successfully applying the chain rule.

Product Rule

When faced with the product of two distinct functions, the product rule is your go-to technique. It helps differentiate functions that are multiplied together, as is needed for calculating the second derivative of our function.

The rule is defined as \(\frac{d}{dx}(v(x)w(x)) = v'(x)w(x) + v(x)w'(x)\). Here, we consider:

• \(v(x)=-4x\) and \(w(x) = e^{-2 x^{2}}\)

• Compute \(v'(x)=-4\)
• Compute \(w'(x)= -4xe^{-2x^{2}}\), which we previously determined using chain rule.

When inserted into the product rule, the expression is \((-4)e^{-2x^{2}} + (-4x)(-4xe^{-2x^{2}})\) which simplifies to form the second derivative.

Derivatives

Derivatives are at the heart of calculus, representing how a function changes at any given point. In our function \(y = e^{-2x^{2}}\), finding derivatives allows us to explore how rapidly \(y\) changes as \(x\) changes.

The first derivative, reached through the chain rule, describes the rate of change of \(y\) itself. It's often called the slope of the tangent to the curve.

The second derivative, found using both the chain and product rules, informs us of \(y\)'s concavity and inflection points, essential for understanding the curve’s overall geometry.