# Thread: use logarthmic differntiation to find dy/dx

1. ## use logarthmic differntiation to find dy/dx

y=[(x+2)*√(1-x²)]/[4x³]

i got as far as

y'/y=ln(x+2)+1/2ln(1-x²)-3ln4x

im not sure if the part is correct since the cubed is only on the x and not the 4.

2. Originally Posted by johnnyc418
y=[(x+2)*√(1-x²)]/[4x³]

i got as far as

y'/y=ln(x+2)+1/2ln(1-x²)-3ln4x

im not sure if the part is correct since the cubed is only on the x and not the 4.
$y = \frac{(x+2) \sqrt{1-x^2}}{4x^3}$

note that $ln(4x^3)$ can be simplified to $ln(4)+ln(x^3) = 2ln(2)+3ln(x)$

$ln(y) = ln(x+2) + \frac{1}{2}ln(1-x^2) - 2ln(2) - 3ln(x)$

$\frac{y'}{y} = \frac{1}{x+2} - \frac{2x}{2(1-x)^2} - \frac{3}{x}$

3. alright so knowing that, would the answer be

y'=([(x+2)*√(1-x²)]/[4x³])*([1/x+2]-[x/1-x²]-3/x]

or is there any simplifying i could do it it. thanks

4. Originally Posted by johnnyc418
alright so knowing that, would the answer be

y'=([(x+2)*√(1-x²)]/[4x³])*([1/x+2]-[x/1-x²]-3/x]

or is there any simplifying i could do it it. thanks
Indeed it would be, you might be able to work on it but I wouldn't bother