Question

Using chain rule:
y=In(px+q)

Answer 1

To differentiate the function \( y = \ln(px + q) \) using the chain rule, we can follow these steps:

1. Identify the outer function and the inner function:
- Outer function: \( u = \ln(v) \), where \( v = px + q \)
- Inner function: \( v = px + q \)

2. Differentiate the outer function with respect to the inner function:
\[
\frac{du}{dv} = \frac{1}{v} = \frac{1}{px + q}
\]

3. Differentiate the inner function with respect to \( x \):
\[
\frac{dv}{dx} = \frac{d}{dx}(px + q) = p
\]

4. Apply the chain rule:
\[
\frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}
\]

5. Combine the results:
\[
\frac{dy}{dx} = \frac{1}{px + q} \cdot p = \frac{p}{px + q}
\]

Therefore, the derivative of the function \( y = \ln(px + q) \) with respect to \( x \) is:
\[
\frac{dy}{dx} = \frac{p}{px + q}
\]