1. Find from the definition, the derivative of a. $$ \ln (a x+b) $$

Question

UpStudy AI · Accepted Answer

To find the derivative of the function $$ f(x) = \ln(ax + b) $$ using the definition of the derivative, we start with the limit definition:

$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$

Substituting $$ f(x) = \ln(ax + b) $$:

$$
f'(x) = \lim_{h \to 0} \frac{\ln(a(x+h) + b) - \ln(ax + b)}{h}
$$

This simplifies to:

$$
f'(x) = \lim_{h \to 0} \frac{\ln(ax + ah + b) - \ln(ax + b)}{h}
$$

Using the property of logarithms that states $$ \ln(A) - \ln(B) = \ln\left(\frac{A}{B}\right) $$:

$$
f'(x) = \lim_{h \to 0} \frac{1}{h} \ln\left(\frac{ax + ah + b}{ax + b}\right)
$$

Now, we can rewrite this as:

$$
f'(x) = \lim_{h \to 0} \frac{1}{h} \ln\left(1 + \frac{ah}{ax + b}\right)
$$

As $$ h \to 0 $$, the term $$ \frac{ah}{ax + b} $$ approaches 0. We can use the fact that $$ \ln(1 + u) \approx u $$ when $$ u $$ is small:

$$
\ln\left(1 + \frac{ah}{ax + b}\right) \approx \frac{ah}{ax + b}
$$

Thus, we have:

$$
f'(x) = \lim_{h \to 0} \frac{1}{h} \cdot \frac{ah}{ax + b} = \lim_{h \to 0} \frac{a}{ax + b}
$$

Since the limit does not depend on $$ h $$, we can evaluate it directly:

$$
f'(x) = \frac{a}{ax + b}
$$

Therefore, the derivative of $$ \ln(ax + b) $$ is:

$$
\boxed{\frac{a}{ax + b}}
$$
The derivative of $$ \ln(ax + b) $$ is $$ \frac{a}{ax + b} $$.

Find from the definition, the derivative of
a.