Question

86-89. Second derivatives Find \(\frac{d^{2} y}{d x^{2}}\) for the following functions. $$y=e^{-2 x^{2}}$$

Short answer

Answer: The second derivative of the function \(y=e^{-2 x^{2}}\) is \(\frac{d^2 y}{d x^2} = -4e^{-2 x^2} + 16x^2 e^{-2 x^2}\).

Step-by-step solution

Differentiate to find the first derivative

We need to find the first derivative, \(\frac{dy}{dx}\), of the function \(y=e^{-2 x^{2}}\). To do this, we will apply the chain rule, which states that the derivative of an outer function multiplied by the derivative of the inner function. Here, \(f(x)=e^x\) is the outer function, and \(g(x)=-2x^2\) is the inner function. So, we have \(y=f(g(x)) = e^{-2x^2}\). The derivative of the outer function, \(f'(x)=e^x\), and the derivative of the inner function, \(g'(x)=-4x\). Now we can apply the chain rule: $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = e^{-2 x^2} \cdot (-4x)$$

Differentiate to find the second derivative

Now that we have the first derivative, we will differentiate it again to find the second derivative, \(\frac{d^2 y}{d x^2}\). We have: \(\frac{dy}{dx} = -4xe^{-2 x^2}\) Now we'll apply the product rule which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. Let \(h(x)=-4x\) and \(i(x)=e^{-2 x^2}.\) Then, the first derivative can be written as a product of two functions: \(\frac{dy}{dx} = h(x)i(x)\) Now, we will find the derivative of each of these functions. The derivative of \(h(x)\) is: $$h'(x)=-4$$ And for the derivative of \(i(x)\), we'll apply the chain rule again (since it involves an inner function \(-2x^2\)): $$i'(x) = e^{-2 x^2} \cdot (-4x) = -4xe^{-2 x^2}$$ Now, we can apply the product rule to find the second derivative: $$\frac{d^2y}{dx^2} = h'(x)i(x) + h(x)i'(x) = -4e^{-2 x^2} + (-4x)(-4xe^{-2 x^2})$$ So, the second derivative of the given function \(y=e^{-2 x^2}\) is: $$\frac{d^2 y}{d x^2} = -4e^{-2 x^2} + 16x^2 e^{-2 x^2}$$

Key concepts

Chain Rule

Understanding the chain rule is crucial when dealing with the differentiation of composite functions. A composite function is created when one function is nested inside another. For example, if we have two functions, say, f(x) and g(x), the composite function would be f(g(x)).

The chain rule tells us how to find the derivative of this composite function. It states that the derivative of f(g(x)) with respect to x is the derivative of f with respect to g(x) multiplied by the derivative of g with respect to x or mathematically expressed as \(\frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\).

Application in the Exercise

In the given exercise, we apply the chain rule to find the first derivative of y=e^{-2x^2}. Here, f(x)=e^x is the outer function and g(x)=-2x^2 is the inner function. By using the chain rule, we differentiate the outer function, substitute the inner function into it, and then multiply by the derivative of the inner function, thus efficiently handling the complexity of a composite function.

Product Rule

When it comes to differentiating the product of two functions, the product rule is indispensable. This rule states that to find the derivative of a product of two differentiable functions, you take the derivative of the first function and multiply it by the second function, and add the product of the first function and the derivative of the second function. The formula is given by \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).

Application in the Exercise

For the second derivative in our exercise, the product rule is employed. We already have the first derivative \(\frac{dy}{dx} = -4xe^{-2 x^2}\), which is a product of \(h(x)=-4x\) and \(i(x)=e^{-2 x^2}\). Using the product rule, we differentiate each function individually and then combine them to find the second derivative. This rule ensures that all parts of the product are accounted for in differentiation.

First Derivative

The first derivative of a function represents the rate at which the function's value changes with respect to its variable. It provides important information about the behavior of the function such as the slope of the tangent line at any point and the identification of critical points, which can correspond to the function's maxima and minima.

Importance of the First Derivative in Second Derivatives

The process of finding the second derivative inherently begins with the first derivative. As in the provided exercise, once the first derivative is obtained through rules appropriate for the function's form, it then serves as the starting point for further differentiation. Understanding the first derivative is essential before attempting to find higher-order derivatives because it sets the foundation for subsequent calculus operations.

Exponential Functions

Exponential functions have the form \(f(x) = a^x\), where \(a\) is a constant, and \(x\) is the variable. They are remarkable for their unique property, where the rate of growth is proportional to the function's current value, making them widely applicable in real-world scenarios such as compound interest calculations, population growth models, and radioactive decay.

Differentiating Exponential Functions

In calculus, an important case of an exponential function is the natural exponential function \(e^x\), which has the special property that its derivative is the same as the original function, \(\frac{d}{dx}e^x = e^x\). In our exercise, y=e^{-2x^2} is a variant of the natural exponential function, having x squared and multiplied by -2 as the exponent. When differentiating such a function, it retains its exponential nature but requires application of the chain rule due to its composite structure. This highlights the exponential function's unique behavior under differentiation, which is often exploited in various calculus problems.