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Answer
The derivative of \( y = \frac{\ln x}{x^5} \) is \( y' = \frac{1 - 5\ln x}{x^6} \).
Solution
1. Express the function:
\[
y=\frac{\ln x}{x^5}
\]
2. To differentiate \( y \), apply the quotient rule:
\[
y'=\frac{v\,u'-u\,v'}{v^2}
\]
where \( u = \ln x \) and \( v = x^5 \).
3. Compute the derivatives individually:
\[
u' = \frac{d}{dx}(\ln x)=\frac{1}{x}
\]
\[
v' = \frac{d}{dx}(x^5)=5x^4
\]
4. Substitute into the quotient rule formula:
\[
y'=\frac{x^5 \cdot \frac{1}{x}-\ln x\cdot 5x^4}{(x^5)^2}
\]
5. Simplify the numerator:
\[
x^5 \cdot \frac{1}{x}=x^4
\]
Hence, the numerator becomes:
\[
x^4-5x^4\ln x=x^4(1-5\ln x)
\]
6. Simplify the denominator:
\[
(x^5)^2=x^{10}
\]
7. Write the derivative:
\[
y'=\frac{x^4(1-5\ln x)}{x^{10}}
\]
8. Cancel the common factor \( x^4 \) in the numerator and the denominator:
\[
y'=\frac{1-5\ln x}{x^6}
\]
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Simplify this solution Beyond the Answer
To differentiate the function \( y = \frac{\ln x}{x^{5}} \), we can use the quotient rule which states that if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \). Here, let \( u = \ln x \) and \( v = x^5 \). First, we compute the derivatives of \( u \) and \( v \): - \( u' = \frac{1}{x} \) - \( v' = 5x^4 \) Now, applying the quotient rule: \[ y' = \frac{\left(\frac{1}{x}\right)(x^5) - (\ln x)(5x^4)}{(x^5)^2} \] Simplifying the numerator: \[ = \frac{x^4 - 5x^4\ln x}{x^{10}} = \frac{x^4(1 - 5\ln x)}{x^{10}} = \frac{1 - 5\ln x}{x^6} \] Thus, the derivative is: \[ y' = \frac{1 - 5\ln x}{x^6} \]
