# Thread: where are these functions differentiable??

1. ## where are these functions differentiable??

a) $\displaystyle f(x) = 1 / sin(x)$ < -- function is differentiable everywhere except 1/2*pi and -1/2*pi...correct??

b) $\displaystyle g(x) = ln(x^2)$...dont know where this one is differentiable

c) $\displaystyle h(x) = f(x) . g(x)$<---- also dont know how to find where this one is differentiable..

2. Originally Posted by b00yeah05
a) $\displaystyle f(x) = 1 / sin(x)$ < -- function is differentiable everywhere except 1/2*pi and -1/2*pi...correct??

b) $\displaystyle g(x) = ln(x^2)$...dont know where this one is differentiable

c) $\displaystyle h(x) = f(x) . g(x)$<---- also dont know how to find where this one is differentiable..

just look where the functions are undefined or have cusps. for (a), it is not differentiable anywhere sin(x) = 0. and that is at infinitely many points (not including 1/2 pi and -1/2 pi), but rather, at k*pi, for k an integer

for (b), the logarithm is undefined at zero, and hence, is not differentiable there

try (c)

if you want to be sure of these answers, try using the limit definition of a derivative, the limit will not exist where the function is not differentiable

3. Originally Posted by b00yeah05
a) $\displaystyle f(x) = 1 / sin(x)$ < -- function is differentiable everywhere except 1/2*pi and -1/2*pi...correct??

b) $\displaystyle g(x) = ln(x^2)$...dont know where this one is differentiable

c) $\displaystyle h(x) = f(x) . g(x)$<---- also dont know how to find where this one is differentiable..

Calculate the first drivative first.

Determine the domain of the gradient function. The exceptions are the x-values where the function f is not differentiable.

$\displaystyle f(x) = \frac1{\sin(x)}~\implies~ f'(x)=-\frac{\cos(x)}{(\sin(x))^2}$

f' is not defined - that means f is not differentiable - if $\displaystyle \sin(x) = 0 ~\implies~ x = k\cdot \pi~,~k \in \mathbb{Z}$

Now it's your turn.

4. c is the hardest one...the first derivative for c i got --- >

$\displaystyle (-cos(x) ln(x^2) / sinx^2) + (2 / sin(x) x)$

i cant simplify it anymore...so you solve the denominator for 0??

5. Originally Posted by b00yeah05
c is the hardest one...the first derivative for c i got --- >

$\displaystyle (-cos(x) ln(x^2) / sinx^2) + (2 / sin(x) x)$

i cant simplify it anymore...so you solve the denominator for 0??
The product in (c) will not be differentiable at any points where either f(x) OR g(x) is not differentiable ...... And you know where both of these aren't differentiable from (a) and (b) .....

6. ahhhhh yes...this one was quite tricky..but thanks for the help mr.fantastic...

7. Originally Posted by mr fantastic
The product in (c) will not be differentiable at any points where either f(x) OR g(x) is not differentiable ...... And you know where both of these aren't differentiable from (a) and (b) .....
I should add that this is not a general rule ...

It works in your particular question but not in general.

The counter-example would be if f(x) = x^2 and g(x) = 1/x, then obviously the product f(x) g(x) IS differentiable at x = 0 .......