Question

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x).\) $$f(x)=x^{2}\left(2+x^{-3}\right)$$

Short answer

Question: Determine the first, second, and third derivatives of the function \(f(x) = x^2(2 + x^{-3})\). Answer: The first derivative of the function is \(f^{\prime}(x) = 4x - x^{-2}\). The second derivative of the function is \(f^{\prime\prime}(x) = 4 + 2x^{-3}\). The third derivative of the function is \(f^{\prime\prime\prime}(x) = -6x^{-4}\).

Step-by-step solution

Differentiate the function using the product rule

To find the first derivative of the function, we can use the product rule \(\frac{d(uv)}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}\). Here, we consider \(u = x^2\) and \(v = 2 + x^{-3}\). Now we differentiate each part: \(\frac{du}{dx} = 2x\) \(\frac{dv}{dx} = −3x^{−4}\) Applying the product rule: $$f^{\prime}(x) = x^{2}\left(-3x^{-4}\right) + \left(2 + x^{-3}\right)(2x)$$

Simplify the first derivative

Now let's simplify the expression we obtained in Step 1: $$f^{\prime}(x) = -3x^{-2} + 4x + 2x^{-2}$$ Combining the terms with the same exponent: $$f^{\prime}(x) = 4x - x^{-2}$$

Differentiate the first derivative using the chain rule

Now find the second derivative using the chain rule \(\frac{d}{dx}\left[f(g(x))\right] = f^{\prime}(g(x)) \cdot g^{\prime}(x)\). We can rewrite \(4x-x^{-2}\) as \(4x-(x^2)^{-1}\). Therefore, \(f(g(x)) = 4x\) and \(g(x) = x^{2}\) for the first term and \(f(g(x)) = -1\) and \(g(x) = x^{2}\) for the second term. Differentiating each term, we get: $$\frac{d}{dx}(4x) = 4$$ $$\frac{d}{dx}(x^2) = 2x$$ Applying the chain rule: $$\frac{d}{dx}(-(x^2)^{-1}) = -1 \cdot (-2x^{-3})$$ So the second derivative is: $$f^{\prime\prime}(x) = 4 + 2x^{-3}$$

Differentiate the second derivative using the chain rule

To find the third derivative, repeat Step 3: $$\frac{d}{dx}(4) = 0$$ $$\frac{d}{dx}(2x^{-3}) = -6x^{-4}$$ So the third derivative is: $$f^{\prime\prime\prime}(x) = -6x^{-4}$$ Thus, we have found the first, second, and third derivatives of the function: $$f^{\prime}(x) = 4x - x^{-2}$$ $$f^{\prime\prime}(x) = 4 + 2x^{-3}$$ $$f^{\prime\prime\prime}(x) = -6x^{-4}$$

Key concepts

Product Rule

The Product Rule is essential for differentiating functions that multiply two expressions together. This rule states that if you have two functions, say \( u(x) \) and \( v(x) \), their derivative can be found using the formula: \[\frac{d}{dx}[u(x) \cdot v(x)] = u(x) \cdot \frac{dv}{dx} + v(x) \cdot \frac{du}{dx}\]Here's a breakdown of how it works:

• First, differentiate \( u(x) \) to get \( \frac{du}{dx} \).
• Then differentiate \( v(x) \) to get \( \frac{dv}{dx} \).
• Finally, apply the rule to obtain the derivative of the product, adding the product of the original \( u(x) \) with \( \frac{dv}{dx} \) to the product of \( v(x) \) with \( \frac{du}{dx} \).

In our example, we took \( u = x^2\) and \(v = 2 + x^{-3}\), leading to: \[f'(x) = x^2(-3x^{-4}) + (2 + x^{-3})2x\] which simplifies to the first derivative \( f'(x) = 4x - x^{-2} \). Understanding each component is crucial for mastering derivatives of complex products.

Chain Rule

The Chain Rule assists you in differentiating compositions of functions, where one function is "inside" another. It's like peeling an onion, where you handle the outer function and then its inner counterpart. The rule is as follows:\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]Let's explore it further:

• First, differentiate the outer function \( f \). Leave the inner function \( g(x) \) as is when doing this step.
• Next, multiply by the derivative of the inner function \( g'(x) \).

In practice, for the derivative \(4x-x^{-2}\), rewritten as \(4x - (x^2)^{-1}\), you see that for the second term, \( f(g(x)) = -(x^2)^{-1} \) where the derivative would involve setting \( g(x) = x^2 \). By applying the Chain Rule, we arrive at the second derivative \( f''(x) = 4 + 2x^{-3} \). This understanding is essential for dealing with nested functions in calculus.

Higher Order Derivatives

Higher Order Derivatives involve finding derivatives beyond the first derivative, indicating how functions change at different levels of application. After the first derivative,

• the second derivative \( f''(x) \) measures the rate of change of the rate of change, i.e., acceleration if you think in terms of motion;
• the third derivative \( f'''(x) \) goes a step further, examining the change in acceleration.

Calculating these requires repeated application of differentiation rules like the Product and Chain Rules. After finding the second derivative \( f''(x) = 4 + 2x^{-3} \) using the Chain Rule, we performed a third differentiation to retrieve \( f'''(x) = -6x^{-4} \).As each derivative provides new insight into the behavior and characteristics of the original function, understanding them can reveal much about the underlying structure of dynamic systems and functions.