What is the derivative of ${(csc x + cot x)}^{- 1}$ $(cscx+cotx)^-1$ ?

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Answer 1 · 2015-05-21T16:47:29

Let's use chain rule here, by naming $u = (csc x + cot x)$ $u=(cscx+cotx)$

Thus, our function is written as $y = u^{- 1}$ $y=u^-1$ .

The chain rule states that

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)(du)/(dx)$

$\frac{d y}{d u} = - u^{- 2}$ $(dy)/(du)=-u^-2$

$\frac{d u}{d x} = - 1 \cdot csc x cot x - 1 \cdot {csc}^{2} x = - csc (cot x + csc x)$ $(du)/(dx)=-1*cscxcotx-1*csc^2x=-csc(cotx+cscx)$

Joining them both:

$(dy)(dx)=(cancel(-)u^-2)/(cancel(-)csc(cotx+cscx))$

Now, substituting $u$ :

$(dy)/(dx)=cancel(cscx+cotx)/(cscxcancel(cscx+cotx))=1/cscx$

What is the derivative of (cscx+cotx)^-1(cscx+cotx)−1(cscx+cotx)^-1?