Find StartFraction dy Over dx EndFraction
  if  yequalsStartFraction 6 x plus 7 Over 42 x Superscript 5 EndFraction
.

To find \(\frac{dy}{dx}\) for the given function \(y = \frac{6x + 7}{42x^5}\), we can use the **quotient rule** for differentiation.

The quotient rule is given as:

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

Here:
- \( u(x) = 6x + 7 \)
- \( v(x) = 42x^5 \)

We now need to find \( u'(x) \) and \( v'(x) \).

1. **Differentiate \( u(x) \):**
   \[
   u'(x) = \frac{d}{dx} (6x + 7) = 6
   \]

2. **Differentiate \( v(x) \):**
   \[
   v'(x) = \frac{d}{dx} (42x^5) = 42 \cdot 5x^4 = 210x^4
   \]

Now, apply the quotient rule:

\[
\frac{dy}{dx} = \frac{(42x^5) \cdot (6) - (6x + 7) \cdot (210x^4)}{(42x^5)^2}
\]

### Simplifying:

- First, expand the numerator:
  \[
  = \frac{42x^5 \cdot 6 - (6x + 7) \cdot 210x^4}{(42x^5)^2}
  \]
  \[
  = \frac{252x^5 - (1260x^5 + 1470x^4)}{(42x^5)^2}
  \]
  \[
  = \frac{252x^5 - 1260x^5 - 1470x^4}{(42x^5)^2}
  \]
  \[
  = \frac{-1008x^5 - 1470x^4}{(42x^5)^2}
  \]

- Now simplify the denominator:
  \[
  (42x^5)^2 = 42^2 \cdot x^{10} = 1764x^{10}
  \]

Thus, the derivative becomes:

\[
\frac{dy}{dx} = \frac{-1008x^5 - 1470x^4}{1764x^{10}}
\]

### Final simplification:

Divide both terms in the numerator by \(1764x^{10}\):

\[
\frac{dy}{dx} = \frac{-1008x^5}{1764x^{10}} - \frac{1470x^4}{1764x^{10}}
\]

Simplifying each term:

\[
\frac{dy}{dx} = \frac{-1008}{1764x^5} - \frac{1470}{1764x^6}
\]
\[
= \frac{-7}{12x^5} - \frac{35}{42x^6}
\]
\[
= \frac{-7}{12x^5} - \frac{5}{6x^6}
\]

Thus, the derivative is:

\[
\frac{dy}{dx} = \frac{-7}{12x^5} - \frac{5}{6x^6}
\]

Would you like further clarification or details?

Here are 5 related questions:
1. How does the quotient rule differ from the product rule in differentiation?
2. Can we simplify the quotient before applying the derivative?
3. How would this derivative change if the powers of \(x\) were different?
4. What is the significance of higher-order derivatives for this function?
5. How would you apply this to solve a real-world problem involving rates of change?

**Tip**: Always simplify expressions before differentiation if possible to reduce complexity!

Learn how to apply the quotient rule to find the derivative of y = (6x + 7) / (42x^5). This problem involves key calculus concepts such as differentiation and the quotient rule, and provides a detailed step-by-step solution. Ideal for undergraduate calculus students.

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