How do you differentiate #arcsin(csc(1-1/x^3)) )# using the chain rule?

Question

Scarlett Allison · Accepted Answer

As a Real valued function this is not differentiable.

This is not differentiable as a Real valued function, since #csc theta in (-oo, -1] uu [1, oo)# and #sin theta in [-1, 1]#, so the only points at which #arcsin(csc(1-1/x^3))# is defined are the discrete points at which: #1-1/x^3 = ((2k+1)pi)/2# That is: #x in { root(3)(1/(1-((2k+1)pi)/2)) : k in ZZ }# If you had asked about #arcsin(csc(1-1/z^3))# then I would look at Complex trigonometric functions.

Ezekiel Alexander · Answer

undefined To differentiate $ \arcsin(\csc(1-\frac{1}{x^3})) $ using the chain rule, we first note that the derivative of $ \arcsin(u) $ with respect to $ u $ is $ \frac{1}{\sqrt{1-u^2}} $.

Next, we differentiate $ \csc(v) $ with respect to $ v $ to get $ -\csc(v) \cot(v) $.

Now, let $ u = \csc(1-\frac{1}{x^3}) $ and $ v = 1-\frac{1}{x^3} $. Then, by the chain rule, the derivative of $ \arcsin(\csc(1-\frac{1}{x^3})) $ with respect to $ x $ is:

$$ \frac{d}{dx} \arcsin(\csc(1-\frac{1}{x^3})) = \frac{d}{dx} \arcsin(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} $$

$$ = \frac{1}{\sqrt{1-\csc^2(1-\frac{1}{x^3})}} \cdot \frac{d}{dx} \csc(1-\frac{1}{x^3}) $$

$$ = \frac{1}{\sqrt{1-\csc^2(v)}} \cdot \frac{d}{dx} \csc(v) $$

$$ = \frac{1}{\sqrt{1-\csc^2(1-\frac{1}{x^3})}} \cdot (-\csc(1-\frac{1}{x^3}) \cot(1-\frac{1}{x^3})) \cdot \frac{d}{dx} (1-\frac{1}{x^3}) $$

$$ = \frac{1}{\sqrt{1-\csc^2(1-\frac{1}{x^3})}} \cdot (-\csc(1-\frac{1}{x^3}) \cot(1-\frac{1}{x^3})) \cdot \frac{3}{x^4} $$