11. [-/1 Points] DETAILS MY NOTES If $$ f(x)+x^{2}[f(x)]^{5}=34 $$ and $$ f(1)=2 $$, find $$ f^{\prime}(1) $$. $$ f^{\prime}(1)=\square $$

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To find $$ f^{\prime}(1) $$ given the equation $$ f(x) + x^{2}[f(x)]^{5} = 34 $$ and the condition $$ f(1) = 2 $$, we will use implicit differentiation.

### Step 1: Differentiate both sides of the equation

We start with the equation:
$$
f(x) + x^{2}[f(x)]^{5} = 34
$$

Differentiating both sides with respect to $$ x $$:
$$
\frac{d}{dx}[f(x)] + \frac{d}{dx}[x^{2}[f(x)]^{5}] = 0
$$

### Step 2: Apply the product rule

For the term $$ x^{2}[f(x)]^{5} $$, we apply the product rule:
$$
\frac{d}{dx}[x^{2}[f(x)]^{5}] = 2x[f(x)]^{5} + x^{2} \cdot 5[f(x)]^{4} \cdot f'(x)
$$

### Step 3: Combine the derivatives

Now we can write the differentiated equation:
$$
f'(x) + \left(2x[f(x)]^{5} + x^{2} \cdot 5[f(x)]^{4} \cdot f'(x)\right) = 0
$$

### Step 4: Solve for $$ f'(x) $$

Rearranging gives:
$$
f'(x) + 2x[f(x)]^{5} + 5x^{2}[f(x)]^{4}f'(x) = 0
$$
$$
f'(x)(1 + 5x^{2}[f(x)]^{4}) = -2x[f(x)]^{5}
$$
$$
f'(x) = \frac{-2x[f(x)]^{5}}{1 + 5x^{2}[f(x)]^{4}}
$$

### Step 5: Evaluate at $$ x = 1 $$

Now we substitute $$ x = 1 $$ and $$ f(1) = 2 $$:
$$
f'(1) = \frac{-2 \cdot 1 \cdot [f(1)]^{5}}{1 + 5 \cdot 1^{2} \cdot [f(1)]^{4}} = \frac{-2 \cdot 1 \cdot 2^{5}}{1 + 5 \cdot 1 \cdot 2^{4}}
$$

Calculating $$ 2^{5} = 32 $$ and $$ 2^{4} = 16 $$:
$$
f'(1) = \frac{-2 \cdot 32}{1 + 5 \cdot 16} = \frac{-64}{1 + 80} = \frac{-64}{81}
$$

### Final Answer

Thus, the value of $$ f^{\prime}(1) $$ is:
$$
f^{\prime}(1) = \frac{-64}{81}
$$
$$ f^{\prime}(1) = -\frac{64}{81} $$

11. [-/1 Points] DETAILS MY NOTES If \( f(x)+x^{2}[f(x)]^{5}=34 \) and \( f(1)=2 \), find \( f^{\prime}(1) \). \( f^{\prime}(1)=\square \)

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