Find f double prime left parenthesis x right parenthesisf′′(x).
​f(x)equals=RootIndex 5 StartRoot left parenthesis x squared plus 1 right parenthesis Superscript 4 EndRoot5x2+14

We are given the function:

\[
f(x) = \sqrt[5]{(x^2 + 1)^4}
\]

We need to find \( f''(x) \), the second derivative of \( f(x) \).

### Step 1: Rewrite the function

First, let's express the function in a more convenient form for differentiation. The fifth root can be rewritten as a power:

\[
f(x) = \left( (x^2 + 1)^4 \right)^{\frac{1}{5}} = (x^2 + 1)^{\frac{4}{5}}
\]

### Step 2: First derivative \( f'(x) \)

To find the first derivative \( f'(x) \), we will use the chain rule. The outer function is \( u^{\frac{4}{5}} \), and the inner function is \( u = x^2 + 1 \).

1. Differentiate the outer function:

\[
\frac{d}{du} \left( u^{\frac{4}{5}} \right) = \frac{4}{5} u^{-\frac{1}{5}}
\]

2. Now, differentiate the inner function \( u = x^2 + 1 \):

\[
\frac{d}{dx} (x^2 + 1) = 2x
\]

Using the chain rule, the first derivative is:

\[
f'(x) = \frac{4}{5} (x^2 + 1)^{-\frac{1}{5}} \cdot 2x = \frac{8x}{5} (x^2 + 1)^{-\frac{1}{5}}
\]

### Step 3: Second derivative \( f''(x) \)

Now, to find the second derivative \( f''(x) \), we need to apply the product rule and the chain rule to the first derivative:

\[
f'(x) = \frac{8x}{5} (x^2 + 1)^{-\frac{1}{5}}
\]

We will differentiate this expression using the product rule, where the two factors are \( \frac{8x}{5} \) and \( (x^2 + 1)^{-\frac{1}{5}} \).

1. Differentiate \( \frac{8x}{5} \):

\[
\frac{d}{dx} \left( \frac{8x}{5} \right) = \frac{8}{5}
\]

2. Differentiate \( (x^2 + 1)^{-\frac{1}{5}} \):

Using the chain rule:

\[
\frac{d}{dx} \left( (x^2 + 1)^{-\frac{1}{5}} \right) = -\frac{1}{5} (x^2 + 1)^{-\frac{6}{5}} \cdot 2x = -\frac{2x}{5} (x^2 + 1)^{-\frac{6}{5}}
\]

Now, applying the product rule:

\[
f''(x) = \frac{8}{5} (x^2 + 1)^{-\frac{1}{5}} + \frac{8x}{5} \cdot \left( -\frac{2x}{5} (x^2 + 1)^{-\frac{6}{5}} \right)
\]

Simplify the second term:

\[
f''(x) = \frac{8}{5} (x^2 + 1)^{-\frac{1}{5}} - \frac{16x^2}{25} (x^2 + 1)^{-\frac{6}{5}}
\]

### Final answer:

\[
f''(x) = \frac{8}{5} (x^2 + 1)^{-\frac{1}{5}} - \frac{16x^2}{25} (x^2 + 1)^{-\frac{6}{5}}
\]

Would you like to explore any steps further or have specific questions?

### 5 Related Questions:
1. How does the chain rule apply to nested functions like \( (x^2 + 1)^4 \)?
2. Can the second derivative be used to find concavity of this function?
3. How would the derivative change if the exponent of \( x^2 + 1 \) were different?
4. What is the importance of the product rule in differentiating composite functions?
5. How can higher-order derivatives help in approximating functions?

### Tip:
When applying the product rule, carefully differentiate each factor while keeping the chain rule in mind for composite expressions.

Learn how to find the second derivative f''(x) of f(x) = (x^2 + 1)^4/5 by applying the chain rule and product rule. This detailed step-by-step solution explains the key calculus concepts involved, suitable for students in Grades 11-12 and early college.

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