Question

a. Derive a formula for the second derivative, \(\frac{d^{2}}{d x^{2}}(f(g(x)))\) b. Use the formula in part (a) to calculate \(\frac{d^{2}}{d x^{2}}\left(\sin \left(3 x^{4}+5 x^{2}+2\right)\right)\)

Short answer

Question: Find the second derivative of the function \(f(x) = \sin(3x^{4}+5x^{2}+2)\). Answer: The second derivative of the function is: \(\frac{d^{2}}{dx^{2}}\left(\sin\left(3x^{4}+5x^{2}+2\right)\right)= -\sin(3x^{4}+5x^{2}+2) \cdot (12x^{3}+10x)^2+\cos(3x^{4}+5x^{2}+2) \cdot (36x^{2}+10)\).

Step-by-step solution

Find the first derivative of f(g(x)) formula

We need to find the first derivative of a composite function, f(g(x)), using the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

Find the second derivative of f(g(x)) formula

Now, we will find the second derivative of f(g(x)) using the chain rule again: \((f(g(x)))'' = \left(f'(g(x)) \cdot g'(x)\right)'\) But we know that we need to differentiate each term in the product, and there are two terms: \(\left(f'(g(x)) \cdot g'(x)\right)'' = (f'(g(x)))' \cdot g'(x) + f'(g(x)) \cdot (g'(x))'\) Applying the chain rule on \((f'(g(x)))'\): \((f'(g(x)))' = f''(g(x)) \cdot g'(x)\) So, the formula for the second derivative is: \((f(g(x)))'' = f''(g(x)) \cdot (g'(x))^2 + f'(g(x)) \cdot g''(x)\)

Given function

We have the given function: \(f(x) = \sin(3x^{4}+5x^{2}+2)\) We need to find the second derivative of this function using the formula derived in Step 2.

Identify f(g(x)) and find f'(g(x)) and f''(g(x))

We can see that the given function can be written as \(f(g(x))\) where \(f(x) = \sin(x)\) and \(g(x) = 3x^{4} + 5x^{2} + 2\). Now let's find the first and second derivatives of f(x) with respect to x: \(f'(x) = \cos(x)\) \(f''(x) = -\sin(x)\)

Find g'(x) and g''(x)

Next, we need to find the first and second derivatives of g(x) with respect to x: \(g'(x) = 12x^{3} + 10x\) \(g''(x) = 36x^{2} + 10\)

Apply the second derivative formula

Finally, we apply the second derivative formula derived in Step 2: \((f(g(x)))'' = f''(g(x)) \cdot (g'(x))^2 + f'(g(x)) \cdot g''(x)\) Substitute the values of f''(g(x)), (g'(x))^2, f'(g(x)), and g''(x) in the formula: \(-\sin(3x^{4}+5x^{2}+2) \cdot (12x^{3}+10x)^2+\cos(3x^{4}+5x^{2}+2) \cdot (36x^{2}+10)\) Hence, the second derivative of the given function is: \(\frac{d^{2}}{dx^{2}}\left(\sin\left(3x^{4}+5x^{2}+2\right)\right)= -\sin(3x^{4}+5x^{2}+2) \cdot (12x^{3}+10x)^2+\cos(3x^{4}+5x^{2}+2) \cdot (36x^{2}+10)\)

Key concepts

Composite Function

A composite function is formed when one function is applied to the results of another function. Imagine you have two functions, say \(f(x)\) and \(g(x)\). A composite function, represented as \(f(g(x))\), implies that you first apply \(g(x)\) and then apply \(f(x)\) to the result of \(g(x)\).

This concept is essential when working with the chain rule in calculus. The chain rule helps in differentiating composite functions by breaking them down into simpler parts.

The idea is that you can calculate the derivative of the outer function (\(f\)) with respect to the inner function (\(g(x)\)), and then multiply it by the derivative of the inner function itself (\(g(x)\)).

Understanding how to handle composite functions efficiently is key to mastering differentiation in calculus.

Second Derivative

The second derivative of a function provides information about the function's curvature or concavity. It is the derivative of the derivative, which means you're looking at how the rate of change itself is changing.

In terms of applications, knowing the second derivative can help in understanding the behavior of a function's graph over different intervals. For instance, a positive second derivative indicates that the graph is concave up (like a cup), while a negative second derivative shows that the graph is concave down.

When finding the second derivative of a composite function, as suggested by the given exercise, you use the derivative of the first derivative formula. By applying the chain rule twice, you can derive the second derivative of functions like \(f(g(x))\), resulting in a formula that combines the derivatives of both \(f\) and \(g\) in different ways.

Mastering the steps to correctly compute the second derivative is crucial for understanding the deeper mechanics of functions.

Differentiation Formulas

Differentiation formulas are the essential tools in calculus for finding how a function changes at any given point.

The most common formulas include the power rule, product rule, quotient rule, and chain rule, each designed for different function types.

• The power rule is typically used for polynomials, making it straightforward to differentiate expressions like \(x^n\).
• The product and quotient rules handle functions that are products or quotients of two simpler functions. They are essential for complex expressions where functions are multiplied or divided.
• The chain rule is specifically used for composite functions, as discussed earlier, and is critical for differentiating functions within functions efficiently.

By leveraging these rules, one can solve a variety of differential calculus problems, including determining the rate at which functions increase or decrease and finding the acceleration represented by second derivatives.

Having a strong grasp of these formulas and knowing when to apply each of them is key to succeeding in calculus.