Find the derivative of the function f(x)= (e^(x^3+4x+2))/((x+1^2))
a) 0         b) ((-(e^(x^3+4x+2))/(x+1)^4)+(e^(x^3+4x+2)(3x^2))/((x+1)^2)             c) (-2(e^(x^3+4x+2))/((x+1^4))+((e^(x^3+4x+2)(3x^2+4))/(x+1)^3               d) (-2(e^(x^3+4x+2)))/(x+1^3)+(e^(x^3+4x+2)(3x^2+4))/(x+1)^2                  e) None of the above

We are tasked with finding the derivative of the function:

\[
f(x) = \frac{e^{x^3 + 4x + 2}}{(x + 1)^2}
\]

This is a quotient, so we will apply the **quotient rule**. The quotient rule states that for two differentiable functions \( u(x) \) and \( v(x) \),

\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}
\]

In our case:
- \( u(x) = e^{x^3 + 4x + 2} \)
- \( v(x) = (x + 1)^2 \)

### Step 1: Differentiate the numerator \( u(x) \)
The derivative of \( u(x) = e^{x^3 + 4x + 2} \) using the chain rule is:

\[
u'(x) = e^{x^3 + 4x + 2} \cdot (3x^2 + 4)
\]

### Step 2: Differentiate the denominator \( v(x) \)
The derivative of \( v(x) = (x + 1)^2 \) is:

\[
v'(x) = 2(x + 1)
\]

### Step 3: Apply the quotient rule
Now, applying the quotient rule:

\[
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}
\]

Substitute the values we computed for \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \):

\[
f'(x) = \frac{e^{x^3 + 4x + 2}(3x^2 + 4)(x + 1)^2 - e^{x^3 + 4x + 2} \cdot 2(x + 1)}{(x + 1)^4}
\]

### Step 4: Simplify
Factor out \( e^{x^3 + 4x + 2} \) from the numerator:

\[
f'(x) = \frac{e^{x^3 + 4x + 2} \left[ (3x^2 + 4)(x + 1)^2 - 2(x + 1) \right]}{(x + 1)^4}
\]

Now simplify the expression inside the brackets:

\[
(3x^2 + 4)(x + 1)^2 = (3x^2 + 4)(x^2 + 2x + 1) = 3x^4 + 6x^3 + 3x^2 + 4x^2 + 8x + 4 = 3x^4 + 6x^3 + 7x^2 + 8x + 4
\]

Thus:

\[
f'(x) = \frac{e^{x^3 + 4x + 2} \left[ (3x^4 + 6x^3 + 7x^2 + 8x + 4) - 2(x + 1) \right]}{(x + 1)^4}
\]

Simplifying further:

\[
f'(x) = \frac{e^{x^3 + 4x + 2} \left( 3x^4 + 6x^3 + 7x^2 + 6x + 2 \right)}{(x + 1)^4}
\]

### Step 5: Match the options
The final result matches option **d**:

\[
f'(x) = \frac{-2e^{x^3 + 4x + 2}}{(x + 1)^3} + \frac{e^{x^3 + 4x + 2}(3x^2 + 4)}{(x + 1)^2}
\]

Thus, the correct answer is **d**.

---

Would you like further details or have any questions? Here are 5 follow-up questions you might explore:

1. What are the conditions under which the quotient rule is applicable?
2. How does the chain rule help in differentiating composite functions like \( e^{x^3 + 4x + 2} \)?
3. Can you derive a similar result for higher powers in the denominator?
4. How would the solution change if the numerator were more complex (e.g., trigonometric functions)?
5. What happens if we differentiate the result again?

**Tip**: Always try to simplify expressions before applying rules like the quotient rule to avoid mistakes.

Find the derivative of the function f(x)= (e^(x^3+4x+2))/((x+1)^2)
a) 0
b) ((-(e^(x^3+4x+2))/(x+1)^4)+(e^(x^3+4x+2)(3x^2))/((x+1)^2)
c) (-2(e^(x^3+4x+2))/((x+1^4))+((e^(x^3+4x+2)(3x^2+4))/(x+1)^3
d) (-2(e^(x^3+4x+2)))/(x+1^3)+(e^(x^3+4x+2)(3x^2+4))/(x+1)^2
e) None of the above

Learn how to find the derivative of f(x) = (e^(x^3+4x+2)) / (x+1)^2 using the quotient rule and chain rule. This step-by-step guide simplifies the process of differentiating complex functions, making it ideal for students tackling advanced calculus problems.

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