What is the derivative of ${log}_{3} x$ $log_3x$ ?

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Answer 1 · 2018-03-31T20:40:01

$\frac{d}{d x} ({log}_{3} x) = \frac{1}{x ln 3}$ $d/dx(log_3x)=1/(xln3)$

We can rewrite ${log}_{3} x$ $log_3x$ as $\frac{ln (x)}{ln (3)}$ $ln(x)/ln(3)$ .

So, we really want $\frac{d}{d x} \frac{ln x}{ln 3}$ $d/dx(lnx)/ln3$ . Knowing that $\frac{d}{d x} ln x = \frac{1}{x}$ $d/dxlnx=1/x$ , we get

$\frac{d}{d x} \frac{ln x}{ln 3} = \frac{1}{x ln 3}$ $d/dx(lnx)/ln3=1/(xln3)$ .

This gives rise to the general differentiation rule

$\frac{d}{d x} {log}_{a} x = \frac{1}{x ln a} .$ $d/dxlog_ax=1/(xlna).$

Answer 2 · 2018-03-31T21:14:02

$\frac{d}{d x} [{log}_{3} (x)] = \frac{1}{ln (3) \cdot x}$ $d/dx[log_3(x)]=1/(ln(3)*x)$

There is a rule here:

$\frac{d}{d x} [{log}_{a} (x)] = \frac{1}{(ln (a) \cdot x)} \cdot \frac{d}{d x} [x]$ $d/dx[log_a(x)]=1/((ln(a)*x))*d/dx[x]$

Therefore:

$\frac{d}{d x} [{log}_{3} (x)] = \frac{1}{ln (3) \cdot x} \cdot \frac{d}{d x} [x]$ $d/dx[log_3(x)]=1/(ln(3)*x)*d/dx[x]$

The derivative of $x$ $x$ is just $1$ $1$ . Therefore:

$\frac{d}{d x} [{log}_{3} (x)] = \frac{1}{ln (3) \cdot x}$ $d/dx[log_3(x)]=1/(ln(3)*x)$

What is the derivative of log3xlog_3x?