Question

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

Short answer

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

Step-by-step solution

Find the first derivative, \(\frac{dy}{dx}\) of \(y = \sin x^2\).

The chain rule states that if a function \(y\) is a composition of two functions \(u(x)\) and \(v(x)\), namely \(y = u(v(x))\), then the derivative of \(y\) with respect to \(x\) is given by $$\frac{dy}{dx} = \frac{du}{dv}\cdot\frac{dv}{dx}$$ In our case, let \(u(v) = \sin v\) and \(v(x) = x^2\). Then, $$\frac{du}{dv} = \cos v$$ and $$\frac{dv}{dx} = 2x$$ Applying the chain rule, $$\frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = \cos v \cdot 2x = \cos (x^2) \cdot 2x$$

Find the second derivative, \(\frac{d^2 y}{d x^2}\), of \(y = \sin x^2\).

Now, we have to find the second derivative by differentiating the first derivative with respect to \(x\). Let \(w(x) = \cos x^2\), then $$\frac{dy}{dx} = 2x\cdot w(x)$$ Applying the product rule, $$\frac{d^2 y}{d x^2} = \frac{d}{dx}(2x\cdot w(x)) = 2\cdot \frac{dw}{dx} \cdot x + 2\cdot w(x)$$ To find \(\frac{dw}{dx}\), we can apply the chain rule to \(w(x) = \cos x^2\). Let \(u(v) = \cos v\) and \(v(x) = x^2\), then $$\frac{dw}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = -\sin v\cdot 2x = -2x\sin(x^2)$$ Substituting this back into the expression for the second derivative, we get $$\frac{d^2 y}{d x^2} = 2(-2x\sin(x^2))\cdot x + 2\cos(x^2) = -4x^2\sin(x^2) + 2\cos(x^2)$$ So, the second derivative of \(y = \sin x^2\) with respect to \(x\) is $$\frac{d^2 y}{d x^2} = -4x^2\sin(x^2) + 2\cos(x^2)$$

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of a composite function. A composite function is created when one function is inside another. Imagine it like layers in a cake, each layer representing a different function. To differentiate these composite functions, we use the chain rule.

The rule states: if you have a function written as \( y = u(v(x)) \) where \( u \) and \( v \) are functions of \( x \), then the derivative \( \frac{dy}{dx} \) is obtained by multiplying the derivative of the outer function \( u \) with respect to \( v \), \( \frac{du}{dv} \), by the derivative of the inner function \( v \) with respect to \( x \), \( \frac{dv}{dx} \).

• Identify the outer function, \( u(v) \).
• Find \( \frac{du}{dv} \).
• Identify the inner function, \( v(x) \).
• Find \( \frac{dv}{dx} \).
• Multiply the results: \( \frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \).

For instance, if we consider the function \( y = \sin x^2 \), we can break it down as \( u(v) = \sin v \) where \( v(x) = x^2 \). This results in \( \frac{du}{dv} = \cos v \) and \( \frac{dv}{dx} = 2x \). Once we multiply these, we get the first derivative \( \frac{dy}{dx} = 2x \cos(x^2) \).

Product Rule

The product rule is another essential principle in calculus, especially when dealing with the differentiation of products of two functions. Imagine having two functions, say \( f(x) \) and \( g(x) \), and these functions are multiplied together to form \( y = f(x) \cdot g(x) \). To find how this product changes with respect to \( x \), the product rule is your go-to strategy.

Here's how the product rule works for differentiation: if \( y = f(x) \cdot g(x) \), the derivative \( \frac{dy}{dx} \) is given by:

• Differentiating \( f(x) \) to get \( \frac{df}{dx} \).
• Differentiating \( g(x) \) to get \( \frac{dg}{dx} \).
• The product rule formula: \( \frac{dy}{dx} = f(x) \cdot \frac{dg}{dx} + g(x) \cdot \frac{df}{dx} \).

In applying the product rule to the second derivative of \( y = \sin x^2 \) starting from \( \frac{dy}{dx} = 2x \cos(x^2) \), we treat \( 2x \) as \( f(x) \) and \( \cos(x^2) \) as \( g(x) \). When we differentiate these, we not only apply the product rule but also the chain rule once again for \( g(x) \). The final result for the second derivative \( \frac{d^2y}{dx^2} \) incorporates these steps and gives \( -4x^2 \sin(x^2) + 2 \cos(x^2) \).

Derivatives

Understanding derivatives is like learning how to gauge the rate of change of a function. Simply put, when you're finding a derivative, you're asking the question: "How does the output of the function change as the input changes?"

Derivatives can be applied to a wide range of functions and reveal critical information about their behavior. Here's how to think about them:

• First Derivative: It gives the slope of the tangent line to the curve at any point. For a function \( y = f(x) \), the first derivative \( \frac{dy}{dx} \) provides insight into how quickly or slowly \( y \) changes with \( x \).
• Second Derivative: It provides information about the concavity of the function. The second derivative \( \frac{d^2y}{dx^2} \) tells us whether the curve is opening upwards or downwards. It's like a higher-level checkpoint on how the rate of change itself is changing.

In our example with \( y = \sin x^2 \), the first derivative \( 2x \cos(x^2) \) indicates initial tendencies, and going further to the second derivative, \( -4x^2 \sin(x^2) + 2 \cos(x^2) \), provides insights into how these tendencies develop over time. Each step helps us better understand the overall behavior of the function, making calculus a powerful tool in understanding change.