Question

find the derivative of y with respect to x.
\frac{ln(19x)}{19x}
\frac{dy}{dx}=square

Explanation:

Step1: Use quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \ln(19x)$ and $v = 19x$.

Step2: Find $u^\prime$

The derivative of $\ln(19x)$ using the chain - rule. Let $t = 19x$, then $\frac{d}{dx}\ln(19x)=\frac{d}{dt}\ln(t)\cdot\frac{dt}{dx}$. Since $\frac{d}{dt}\ln(t)=\frac{1}{t}$ and $\frac{dt}{dx}=19$, so $u^\prime=\frac{1}{19x}\cdot19=\frac{1}{x}$. And $v^\prime = 19$.

Step3: Substitute into quotient - rule

$y^\prime=\frac{\frac{1}{x}\cdot19x-\ln(19x)\cdot19}{(19x)^{2}}$.
Simplify the numerator: $\frac{1}{x}\cdot19x-\ln(19x)\cdot19 = 19 - 19\ln(19x)$.
So $y^\prime=\frac{19 - 19\ln(19x)}{(19x)^{2}}=\frac{1-\ln(19x)}{19x^{2}}$.

Answer:

$\frac{1-\ln(19x)}{19x^{2}}$