#f(x)# composed of two functions #cscx " "and" "Ln x#.
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So, Differentiating #f(x)# is determined by applying chain rule.
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Let #u(x)=ln(1/x)" " then " "f(x)= csc(u(x))#
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#d/dx(f(x))=d/dx(cscx(u(x)))#
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#color(blue)(d/dx(f(x))=u'(x) xx csc'(u(x)))#
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Let us compute #color(blue)(u'(x))#
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#u'(x)=(color(red)((1/x)'))/(1/x)#
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What is the derivative of #1/x=?#
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The differiation of #1/x# is determined by applying the quotient rule .
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#color(red)((1/x)')=((1)'xxx-(x)'xx1)/x^2= (0-1)/x^2=color(red)(-1/x^2)#
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#color(blue)(u'(x))=(-1/x^2)/(1/x)=color(blue)(-1/x)#
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Knowing that the derivative of #cscx" "#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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#color(blue)(d/dx(f(x))=u'(x) xx csc'(u(x)))#
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#d/dx(f(x))=(-1/x)xx(-cotln(1/x)xxcscln(1/x))#
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Therefore,
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#d/dx(f(x))=1/xcotln(1/x)cscln(1/x)#
