Pope Mullins
07/23/2023 · Junior High School
Use the general power rule twice to differentiate. \[ \begin{aligned} f(x)=\sqrt[7]{\left(3 x^{4}-7\right)^{4}-x} \\ f^{\prime}(x)=\end{aligned} \]
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To differentiate the function
\[
f(x) = \sqrt[7]{(3x^4 - 7)^4 - x},
\]
we will use the general power rule and the chain rule.
First, we can rewrite the function in a more convenient form for differentiation:
\[
f(x) = \left((3x^4 - 7)^4 - x\right)^{1/7}.
\]
Now, we will apply the chain rule. Let
\[
u = (3x^4 - 7)^4 - x.
\]
Then, we can express \( f(x) \) as:
\[
f(x) = u^{1/7}.
\]
Now, we differentiate \( f(x) \) with respect to \( x \):
\[
f'(x) = \frac{1}{7} u^{-6/7} \cdot u'.
\]
Next, we need to find \( u' \). We have:
\[
u = (3x^4 - 7)^4 - x.
\]
Differentiating \( u \):
\[
u' = \frac{d}{dx}((3x^4 - 7)^4) - \frac{d}{dx}(x).
\]
Using the chain rule on the first term:
\[
\frac{d}{dx}((3x^4 - 7)^4) = 4(3x^4 - 7)^3 \cdot \frac{d}{dx}(3x^4 - 7).
\]
Now, we differentiate \( 3x^4 - 7 \):
\[
\frac{d}{dx}(3x^4 - 7) = 12x^3.
\]
Putting it all together, we have:
\[
\frac{d}{dx}((3x^4 - 7)^4) = 4(3x^4 - 7)^3 \cdot 12x^3 = 48x^3(3x^4 - 7)^3.
\]
Now, we differentiate \( -x \):
\[
\frac{d}{dx}(-x) = -1.
\]
Thus, we can write \( u' \) as:
\[
u' = 48x^3(3x^4 - 7)^3 - 1.
\]
Now substituting \( u' \) back into the expression for \( f'(x) \):
\[
f'(x) = \frac{1}{7} u^{-6/7} \cdot (48x^3(3x^4 - 7)^3 - 1).
\]
Finally, substituting back \( u = (3x^4 - 7)^4 - x \):
\[
f'(x) = \frac{1}{7} \left((3x^4 - 7)^4 - x\right)^{-6/7} \cdot (48x^3(3x^4 - 7)^3 - 1).
\]
Thus, the derivative \( f'(x) \) is:
\[
\boxed{f'(x) = \frac{1}{7} \left((3x^4 - 7)^4 - x\right)^{-6/7} \cdot (48x^3(3x^4 - 7)^3 - 1)}.
\]
Quick Answer
The derivative \( f'(x) \) is:
\[
f'(x) = \frac{1}{7} \left((3x^4 - 7)^4 - x\right)^{-6/7} \cdot (48x^3(3x^4 - 7)^3 - 1).
\]
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