Question

Find the derivative of y=lnsqrt(x^2-3)

Answer 1

start with
\[y=\frac{1}{2}\ln(x^2-3)\] and then it will be much easier

Answer 2

using
\[\frac{d}{dx}[\ln(f(x))]=\frac{f'(x)}{f(x)}\]

Answer 3

you can just about do it in your head.
try it

Answer 4

Thanks! Yup I got there. If I let u = x^2 -3, does dy/du = 1/2u?

Answer 5

oh, we haven't learnt that formula yet.

Answer 6

what "formula"?

Answer 7

this
ddx[ln(f(x))]=f′(x)f(x) formula

Answer 8

ignore the 1/2 out front, get
\[\frac{2x}{x^2-3}\] instantly, then divide by 2 and get
\[\frac{x}{x^2-3}\]

Answer 9

Use the chain rule.
\[\frac{d(ln\sqrt{x^2 - 3})}{dx} = \frac{1}{2}\frac{d(ln\sqrt{x^2-3})}{dx}\]

now, derivative of ln(x) is 1/x.
but here, the argument of the log is (x^2 - 3). So, use chain rule.

Answer 10

it is not a "formula" it the chain rule

Answer 11

the derivative of log of x is one over x
the derivative of log of something is one over something times the derivative of something
that is the chain rule in action
more succinctly written as
\[\frac{d}{dx}[\ln(f(x))]=\frac{f'(x)}{f(x)}\]

Answer 12

@apoorvk think there is a typo there.
square root should be gone after the equal sign

Answer 13

Oops sorry! thanks @satellite73 , yes the sqrt would be gone after the equal sign. should read:
\[\frac{d(ln\sqrt{x^2 - 3})}{dx} = \frac{1}{2}\frac{d~ln{(x^2-3})}{dx}\]

Answer 14

Hmm.. ok I understand the chain rule but when I used to u to sub x^2 -3, y = 0.5lnu, d(0.5linu)/du = 1/2u ... which isn't the top number that you can get from doing f'(x)/f(x)

Answer 15

OH I get it now! Thank you both!!