differentiate

• May 8th 2011, 05:52 AM
Jon123
differentiate
Hi there!

Can someone please explain to me the reason for,

y = In[In(sin x)]

does it exist?
and the answer cannot be
dy/dx = cot x / In (sin x) because it is apparently it is incorrect.

would the anwer be: dy/dx = In (sin x) / In (sin x)
but if it is, it does not make sense when using the formula of,
d/dx (In f(x)) = f ' (x) / f (x )

So, would the answer then be, -1 < sin x < 1
or is the answer, 0 < sin x < 1

and also with y = In[In(cos x)]
does not exist due to (In cos x) always being negative.

Thanks heaps if you get this, true champ!!!!!
• May 8th 2011, 06:09 AM
e^(i*pi)
FYI the natural log is written as ln(x) - that's a lower case L.

Looking at the domain of the original equation it would appear that no values satisfy the equation since the argument must be positive.

The range of sin(x) is $-1 \leq \sin(x) \leq 1$. We can immediately discard the negative values (and 0) of sin(x) to satisfy the inner bracket. For $0 \leq x \leq 1$ then $\ln(x) \leq 0$ and as mentioned above the argument of a log must be positive so all values of x lie outside the domain of y.

It would then follow that if the original function cannot be satisfied then there must be no derivative

Hope this makes sense and for what it's worth I get the same value for the derivative as you: $\dfrac{dy}{dx} = \dfrac{cot(x)}{\ln(\sin(x))}$