Question

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

Short answer

Answer: The given function is \(y = \sin{x^2}\), and we are trying to find its second derivative. 2. How do we obtain the first derivative? Answer: We use the chain rule to find the first derivative, which is \(\frac{dy}{dx} = \cos(x^2) \cdot 2x\). 3. What is the second derivative of the function? Answer: The second derivative of the function is \(\frac{d^2y}{dx^2} = -4x^2\sin(x^2) + 2\cos(x^2)\).

Step-by-step solution

Find the first derivative

To find the first derivative, we will use the chain rule. The chain rule states that if we have a function \(y = f(g(x))\), then the derivative of \(y\) with respect to \(x\) is: $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$$ In our case, \(f(x) = \sin{x}\) and \(g(x) = x^2\). So, \(f'(x) = \cos{x}\) and \(g'(x) = 2x\). Now apply the chain rule: $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = \cos(x^2) \cdot 2x$$

Find the second derivative

To find the second derivative, we will differentiate the first derivative with respect to \(x\). We should apply the chain rule again and use product rule if necessary. The product rule states that if we have a function \(y = u(x)v(x)\), then the derivative of \(y\) with respect to \(x\) is: $$\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$$ In our case, \(u(x) = \cos (x^2)\) and \(v(x) = 2x\). So, \(u'(x) = -2x\sin(x^2)\) (again applying the chain rule) and \(v'(x) = 2\). Now apply the product rule: $$\frac{d^2y}{dx^2} = u'(x)v(x) + u(x)v'(x) = -2x\sin(x^2) \cdot 2x + \cos(x^2) \cdot 2$$ Combining the two terms, we get: $$\frac{d^2y}{dx^2} = -4x^2\sin(x^2) + 2\cos(x^2)$$ So, the seond derivative of the function \(y = \sin{x^2}\) is: $$\frac{d^2y}{dx^2} = -4x^2\sin(x^2) + 2\cos(x^2)$$

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. A composite function is when one function is nested within another. This can be visualized as peeling an onion where each layer represents a function we need to differentiate.

To apply the chain rule, suppose you have a function expressed as the composition of two functions, such as \(y = f(g(x))\). The derivative of this function is not as straightforward because it involves more than one part. The chain rule helps by allowing you to differentiate each part step by step.

Here's how it works:

• Differentiating the outer function: First, find the derivative of the outer function \(f\), but leave the inside function \(g(x)\) untouched. This derivative is \(f'(g(x))\).
• Differentiating the inner function: Next, find the derivative of the inside function \(g(x)\), which is \(g'(x)\).
• The final derivative: Multiply these two derivatives together to get the derivative of the composite function: \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).

In our original exercise using the function \(y = \sin{x^2}\), the chain rule helps us find its derivative. We treat \(y\) as a composite function where \(\sin{x}\) is the outer layer and \(x^2\) is the inner layer. Applying the chain rule, we first find the derivative of \(\sin{x}\) to be \(\cos{x}\) while keeping \(x^2\) intact, giving us the expression \(\cos(x^2)\). Then, we find the derivative of \(x^2\) which is \(2x\), giving us \(\frac{dy}{dx} = \cos(x^2) \cdot 2x\).

This combined result gives us the rate of change of the composite function \(\sin{x^2}\) with respect to \(x\).

Product Rule

The product rule is an essential technique in calculus when you need to find the derivative of a product of two functions.

When you have a function expressed as a product such as \(y = u(x)v(x)\), finding its derivative involves more than straightforward differentiation, due to the involvement of two separate parts that are multiplied together.

The product rule formula is:

• Differentiating \(y = u(x)v(x)\), gives \(\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)\).

This formula essentially breaks down the task into finding the derivatives of the two functions separately and then combining them.

In the original exercise, when finding the second derivative, the first derivative was \(\frac{dy}{dx} = \cos(x^2) \cdot 2x\). Here, we treat \(\cos(x^2)\) as \(u(x)\) and \(2x\) as \(v(x)\).

Using the product rule we need:

• \(u'(x)\): Differentiate \(u(x) = \cos(x^2)\) using the chain rule again, giving \(-2x\sin(x^2)\).
• \(v'(x)\): Differentiate \(v(x) = 2x\), giving \(2\).

Finally, plug these into the product rule formula: \(\frac{d^2y}{dx^2} = \left(-2x\sin(x^2)\right)\cdot 2x + \cos(x^2)\cdot 2\).

Solving this step-by-step gives us the second derivative, considering both the changes in \(u(x)\) and \(v(x)\). This is crucial for understanding how two functions interacting with each other change as a whole.

Second Derivative

The second derivative of a function provides insight into the curvature and concavity of the original function. It tells us how the rate of change of a function, already described by the first derivative, itself changes. In simpler terms, while the first derivative gives the slope of the tangent line to the curve at any point, the second derivative gives information about how this slope is changing.

Key aspects of the second derivative include:

• Concavity: If the second derivative \(\frac{d^2y}{dx^2}\) is positive at a point, the function is concave up at that point, resembling a cup shape. If it's negative, the function is concave down, like a cap.
• Points of inflection: A point where the second derivative changes sign (from positive to negative or vice versa) is called a point of inflection. Here the concavity of the function changes.

In the original exercise, after using the chain and product rules, the second derivative of \(y = \sin{x^2}\) was calculated to be \(-4x^2\sin(x^2) + 2\cos(x^2)\). This expression can be used to analyze the concavity of the function \(\sin{x^2}\) with respect to \(x\).

Understanding the behavior through the second derivative is essential, especially in fields like physics and engineering, where knowing not just the rate of change, but how that rate is evolving, is crucial in system dynamics. The second derivative plays a vital role in these deeper insights into the changing conditions and trends of functions.