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Step-by-step Solution

Find the derivative of $\left(\frac{\cos\left(x\right)-1}{\sin\left(x\right)}\right)^{\left(\frac{1}{2}\right)}$

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Answer

$\frac{1}{2}\left(\frac{\cos\left(x\right)-1}{\sin\left(x\right)}\right)^{-\frac{1}{2}}\left(\frac{-\sin\left(x\right)^2-\left(\cos\left(x\right)-1\right)\cos\left(x\right)}{\sin\left(x\right)^2}\right)$

Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\left(\frac{\cos\left(x\right)-1}{\sin\left(x\right)}\right)^{\frac{1}{2}}\right)$
1

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{1}{2}\left(\frac{\cos\left(x\right)-1}{\sin\left(x\right)}\right)^{-\frac{1}{2}}\cdot\frac{d}{dx}\left(\frac{\cos\left(x\right)-1}{\sin\left(x\right)}\right)$
2

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{1}{2}\left(\frac{\cos\left(x\right)-1}{\sin\left(x\right)}\right)^{-\frac{1}{2}}\left(\frac{\frac{d}{dx}\left(\cos\left(x\right)-1\right)\sin\left(x\right)-\left(\cos\left(x\right)-1\right)\frac{d}{dx}\left(\sin\left(x\right)\right)}{\sin\left(x\right)^2}\right)$

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Answer

$\frac{1}{2}\left(\frac{\cos\left(x\right)-1}{\sin\left(x\right)}\right)^{-\frac{1}{2}}\left(\frac{-\sin\left(x\right)^2-\left(\cos\left(x\right)-1\right)\cos\left(x\right)}{\sin\left(x\right)^2}\right)$
$\frac{d}{dx}\left(\left(\frac{\cos\left(x\right)-1}{\sin\left(x\right)}\right)^{\frac{1}{2}}\right)$

Main topic:

Differential calculus

Time to solve it:

~ 1.32 seconds

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