Question

Higher-order derivatives Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) for the following functions. $$f(x)=5 x^{4}+10 x^{3}+3 x+6$$

Short answer

Question: Find the first, second, and third derivatives of the function \(f(x) = 5x^4 + 10x^3 + 3x + 6\). Answer: The derivatives are: \(f'(x) = 20x^3 + 30x^2 + 3\) \(f''(x) = 60x^2 + 60x\) \(f'''(x) = 120x + 60\)

Step-by-step solution

Find the first derivative, \(f'(x)\)

To find the first derivative, apply the power rule of differentiation to each term in the function \(f(x)\). The power rule states that if \(f(x) = ax^n\), then \(f'(x) = nax^{n-1}\). So, for \(f(x) = 5x^4 + 10x^3 + 3x + 6\), \(f'(x) = \frac{d}{dx}(5x^4) + \frac{d}{dx}(10x^3) + \frac{d}{dx}(3x) + \frac{d}{dx}(6)\). Now find the derivatives for each term: \(\frac{d}{dx}(5x^4) = 20x^3\), using the power rule with \(a=5\) and \(n=4\). \(\frac{d}{dx}(10x^3) = 30x^2\), using the power rule with \(a=10\) and \(n=3\). \(\frac{d}{dx}(3x) = 3\), using the power rule with \(a=3\) and \(n=1\). \(\frac{d}{dx}(6) = 0\), as the derivative of constant is always 0. Now, combine all these derivatives to find \(f'(x)\): \(f'(x) = 20x^3 + 30x^2 + 3\).

Find the second derivative, \(f''(x)\)

To find the second derivative, apply the power rule of differentiation again to each term of the first derivative, \(f'(x) = 20x^3 + 30x^2 + 3\): \(f''(x) = \frac{d}{dx}(20x^3) + \frac{d}{dx}(30x^2) + \frac{d}{dx}(3)\). Now find the derivatives for each term: \(\frac{d}{dx}(20x^3) = 60x^2\), using the power rule with \(a=20\) and \(n=3\). \(\frac{d}{dx}(30x^2) = 60x\), using the power rule with \(a=30\) and \(n=2\). \(\frac{d}{dx}(3) = 0\), as the derivative of a constant is always 0. Now, combine all these derivatives to find \(f''(x)\): \(f''(x) = 60x^2 + 60x\).

Find the third derivative, \(f'''(x)\)

Finally, to find the third derivative, apply the power rule of differentiation once more to each term of the second derivative \(f''(x) = 60x^2 + 60x\): \(f'''(x) = \frac{d}{dx}(60x^2) + \frac{d}{dx}(60x)\). Now find the derivatives for each term: \(\frac{d}{dx}(60x^2) = 120x\), using the power rule with \(a=60\) and \(n=2\). \(\frac{d}{dx}(60x) = 60\), using the power rule with \(a=60\) and \(n=1\). Now, combine all these derivatives to find \(f'''(x)\): \(f'''(x) = 120x + 60\).

Final Answer:

For the given function \(f(x)=5 x^{4}+10 x^{3}+3 x+6\), the following derivatives are: \(f'(x) = 20x^3 + 30x^2 + 3\) \(f''(x) = 60x^2 + 60x\) \(f'''(x) = 120x + 60\)

Key concepts

First Derivative

Taking the first derivative of a function is like discovering its initial step in change. The first derivative, noted as \( f'(x) \), gives us a function that describes how our original function \( f(x) \) evolves with small changes in \( x \). It represents the slope or the rate of change of the function at any point on its curve.

To find the first derivative, we can use the power rule. This rule is useful for differentiating terms like \( ax^n \), where \( a \) and \( n \) are constants. The power rule tells us to multiply the power \( n \) by the coefficient \( a \) and reduce the power by one:

\[ \frac{d}{dx}(ax^n) = nax^{n-1} \]

Applying this to \( f(x) = 5x^4 + 10x^3 + 3x + 6 \):

• \( 5x^4 \) becomes \( 20x^3 \)
• \( 10x^3 \) becomes \( 30x^2 \)
• \( 3x \) becomes \( 3 \)
• The constant \( 6 \) becomes zero, as constants have no slope.

Thus, the first derivative is \( f'(x) = 20x^3 + 30x^2 + 3 \). This shows us how the original function is increasing or decreasing at various points.

Second Derivative

The second derivative, represented as \( f''(x) \), is like taking one more step into the future of a function's behavior. It provides us with information on the concavity of the function, indicating whether the graph is curving upwards or downwards.

To find the second derivative, we differentiate the first derivative \( f'(x) \). We apply the power rule again:

From our first derivative \( f'(x) = 20x^3 + 30x^2 + 3 \), we differentiate term by term:

• \( 20x^3 \) becomes \( 60x^2 \)
• \( 30x^2 \) becomes \( 60x \)
• \( 3 \) becomes \( 0 \), as the derivative of a constant is zero.

Combining these, we get \( f''(x) = 60x^2 + 60x \).

This second derivative tells us how the slope itself is changing. If the second derivative is positive, the graph is concave up (like a happy face). If it's negative, the graph is concave down (like a frown). Understanding concavity helps anticipate whether a function might have a local maximum or minimum.

Third Derivative

By the time we reach the third derivative, represented as \( f'''(x) \), we are delving into the realm of understanding the rate of change of the rate of change! This can relate to the physical concept of "jerk" in physics, which is the rate at which acceleration itself changes.

We obtain the third derivative by differentiating the second derivative \( f''(x) \) with respect to \( x \). Applying the power rule to the second derivative \( f''(x) = 60x^2 + 60x \):

• \( 60x^2 \) becomes \( 120x \)
• \( 60x \) becomes \( 60 \)

So, the third derivative is \( f'''(x) = 120x + 60 \).

This third derivative can provide additional understanding in complex problems, such as those in engineering or physics, where knowing how the acceleration itself changes can be crucial. In mathematical terms, though, it often confirms and supports our analysis of functions in broad strokes.