When do you use a **Logarithm** when taking a derivative?
1 Answer
See below.
Explanation:
Take the example of
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
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
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!