Question

When do you use a **Logarithm** when taking a derivative?

Answer 1

See below.

Explanation:

Take the example of `f(x) = ((2 + x)^5(x - 3)^3)/(x + 1)^4`

Differentiating this using the chain/quotient rules would be long and tedious. Therefore, we can use something called logarithmic differentiation.

We start by taking the natural logarithm of both sides.

`lny = ln(((2 + x)^5(x -3)^3)/(x + 1)^4)`

`lny= ln(2 + x)^5 + ln(x- 3)^3 - ln(x +1)^4`

`lny = 5ln(2 + x) + 3ln(x -3) - 4ln(x +1)`

We now differentiate both sides with respect to `x`.

`1/y(dy/dx) = 5/(2 + x) + 3/(x -3) - 4/(x+ 1)`

`dy/dx = y(5/(2 + x) + 3/(x- 3) - 4/(x +1))`

`dy/dx = ((2 +x)^5(x - 3)^3)/(x + 1)^4(5/(2 + x) + 3/(x -3) -4/(x+ 1))`

This is our answer. We can also use it to simplify trig derivatives. Take the example of `g(x) = sec(x)`. Taking the natural logarithm of both sides, we get

`lny = ln(sec(x))`

`lny = ln(1/cos(x))`

`lny = ln(1) - ln(cos(x))`

Now by implicit differentiation on the left and chain rule on the right we have.

`1/y(dy/dx) = 0 - (-sinx) * 1/cosx`

`1/y(dy/dx) = tanx`

`dy/dx =ytanx`

`dy/dx= secxtanx`

Hopefully this helps!