# How do you differentiate #arcsin(sqrt(sin^2(1/x) )# using the chain rule?

##### 2 Answers

Initially, I said I would not use the chain rule. But, see below. (Warning, this looks more complicated to me than using the chain rule turned out to be.)

#### Explanation:

Use

#= {(arcsin(sin(1/x)),"if",sin(1/x) >= 0),(arcsin(-sin(1/x)),"if",sin(1/x) < 0) :}#

#= {(arcsin(sin(1/x)),"if",sin(1/x) >= 0),(arcsin(sin(-1/x)),"if",sin(1/x) < 0) :}#

Now we have to find

**Positive sine**

For every real number

So, there is exactly one integer

If

If

**Negative sine**

For every real number

So, there is exactly one integer

If

And,

If

And,

**Writing #f(x)#**

**Differentiate each branch**

We'll delete the joints, as the derivative will fail to exist if

We might try to clarify by writing

Finally observe that at the joints, the left and right derivatives are not equal, so the derivative does not exist at those points.

Using the chain rule, I get

#### Explanation:

So we get

# = 1/sqrt(1-sin^2(1/x)) [1/(2sqrt(sin^2(1/x))) d/dx(sin^2(1/x))]#

# = 1/(2sqrt(sin^2(1/x))(2sqrt(1-sin^2(1/x))))[2sin(1/x)cos(1/x) (-1/x^2)]#

# = (-sin(1/x)cos(1/x))/(x^2sqrt(sin^2(1/x))sqrt(1-sin^2(1/x)))#

Because

# = (-sin(1/x)cos(1/x))/(x^2sqrt(sin^2(1/x))sqrt(cos^2(1/x)))# .

Or, using

# = (-sin(1/x)cos(1/x))/(x^2abs(sin(1/x))abs(cos^2(1/x)))# .

Those familiar with the fact that

the result might be clearest written as:

In other words, the derivative at

and

if either