# How do you differentiate f(x)=csc(ln(1/x))  using the chain rule?

Nov 22, 2016

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{1}{x} \cot \ln \left(\frac{1}{x}\right) \csc \ln \left(\frac{1}{x}\right)$

#### Explanation:

$f \left(x\right)$ composed of two functions $\csc x \text{ "and" } L n x$.
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So, Differentiating $f \left(x\right)$ is determined by applying chain rule.
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Let $u \left(x\right) = \ln \left(\frac{1}{x}\right) \text{ " then " } f \left(x\right) = \csc \left(u \left(x\right)\right)$
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$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(\csc x \left(u \left(x\right)\right)\right)$
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$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = u ' \left(x\right) \times \csc ' \left(u \left(x\right)\right)}$
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Let us compute $\textcolor{b l u e}{u ' \left(x\right)}$
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$u ' \left(x\right) = \frac{\textcolor{red}{\left(\frac{1}{x}\right) '}}{\frac{1}{x}}$
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What is the derivative of 1/x=?
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The differiation of $\frac{1}{x}$ is determined by applying the quotient rule .
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$\textcolor{red}{\left(\frac{1}{x}\right) '} = \frac{\left(1\right) ' \times x - \left(x\right) ' \times 1}{x} ^ 2 = \frac{0 - 1}{x} ^ 2 = \textcolor{red}{- \frac{1}{x} ^ 2}$
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$\textcolor{b l u e}{u ' \left(x\right)} = \frac{- \frac{1}{x} ^ 2}{\frac{1}{x}} = \textcolor{b l u e}{- \frac{1}{x}}$
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Knowing that the derivative of $\csc x \text{ }$is :
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color(blue)((cscx)'=-cotxcscx
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Let us compute color(blue)(csc'(u(x))
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color(blue)(csc'(u(x)))=-cot(u(x))xxcsc(u(x))=color(blue)(-cotln(1/x)xxcscln(1/x)
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$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = u ' \left(x\right) \times \csc ' \left(u \left(x\right)\right)}$
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$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \left(- \frac{1}{x}\right) \times \left(- \cot \ln \left(\frac{1}{x}\right) \times \csc \ln \left(\frac{1}{x}\right)\right)$
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Therefore,
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$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{1}{x} \cot \ln \left(\frac{1}{x}\right) \csc \ln \left(\frac{1}{x}\right)$