Tyrell Singleton

2023-02-26

How to find the derivative of $y=\mathrm{ln}\left(5x\right)$?

Kylie Woodward

Beginner2023-02-27Added 6 answers

Suppose, $y=\mathrm{ln}\left(b\left(x\right)\right)$

Thus using Chain Rule,

$y\prime =\frac{1}{b\left(x\right)}\cdot \left(b\left(x\right)\right)\prime$

Similarly following for the above function yields,

$y\prime =\frac{1}{5x}\cdot \left(5x\right)\prime$

$y\prime =\frac{1}{5x}\cdot 5$

Hence, $y\prime =\frac{1}{x}$

Thus using Chain Rule,

$y\prime =\frac{1}{b\left(x\right)}\cdot \left(b\left(x\right)\right)\prime$

Similarly following for the above function yields,

$y\prime =\frac{1}{5x}\cdot \left(5x\right)\prime$

$y\prime =\frac{1}{5x}\cdot 5$

Hence, $y\prime =\frac{1}{x}$

Rodney Hoover

Beginner2023-02-28Added 4 answers

A student comfortable with the natural logarithm function and its properties might think of this:

One could reason as follows:

$y=\mathrm{ln}\left(5x\right)=\mathrm{ln}\left(5\right)+\mathrm{ln}\left(x\right)$.

But ln(5) is a constant, so its derivative is 0.

Hence, $\frac{dy}{dx}=\frac{d}{dx}(\mathrm{ln}5+\mathrm{ln}x)=\frac{d}{dx}\left(\mathrm{ln}x\right)=\frac{1}{x}$.

One could reason as follows:

$y=\mathrm{ln}\left(5x\right)=\mathrm{ln}\left(5\right)+\mathrm{ln}\left(x\right)$.

But ln(5) is a constant, so its derivative is 0.

Hence, $\frac{dy}{dx}=\frac{d}{dx}(\mathrm{ln}5+\mathrm{ln}x)=\frac{d}{dx}\left(\mathrm{ln}x\right)=\frac{1}{x}$.