Question

Log(2x+3) derivative?

Answer 1

`color(blue)(2/((2x+3)ln(10)))`

Explanation:

I am assuming this is a base 10 logarithm:

I order to differentiate this, we need to use the change of base formula so as to represent it as base `bbe` logarithms.

`log_ba=lna/lnb`

`:.`

`log(2x+3)=ln(2x+3)/ln10`

Expressing this as:

`ln(2x+3)*1/ln(10)`

`dy/dx(ln(2x+3)*1/ln(10))=1/ln(10)dy/dxln(2x+3)`

Now:

`y=ln(2x+3)<=>e^y=2x+3`

Differentiating implicitly:

`e^ydy/dx=2`

`dy/dx=2/e^y`

From above `e^y=2x+3`

`:.`

`dy/dx=2/(2x+3)`

We now multiply this by `1/ln(10)` which we factored out earlier:

`2/(2x+3)*1/ln(10)=color(blue)(2/((2x+3)ln(10)))`