Differentiate $f ( x ) = frac { ln left( x ^ { 2 } cos ( x ) right) } { sqrt { 1 - x ^ { 2 } } }$












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could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for me to show my own working of the problem.






calculus







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  • $begingroup$
    Read about the chain rule
    $endgroup$
    – Shubham Johri
    Jan 14 at 8:33


















1












$begingroup$


could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for me to show my own working of the problem.






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Read about the chain rule
    $endgroup$
    – Shubham Johri
    Jan 14 at 8:33
















1












1








1





$begingroup$


could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for me to show my own working of the problem.






calculus







share|cite|improve this question









$endgroup$




could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for me to show my own working of the problem.






calculus




calculus






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asked Jan 14 at 8:30









Tariro ManyikaTariro Manyika

619




619












  • $begingroup$
    Read about the chain rule
    $endgroup$
    – Shubham Johri
    Jan 14 at 8:33




















  • $begingroup$
    Read about the chain rule
    $endgroup$
    – Shubham Johri
    Jan 14 at 8:33


















$begingroup$
Read about the chain rule
$endgroup$
– Shubham Johri
Jan 14 at 8:33






$begingroup$
Read about the chain rule
$endgroup$
– Shubham Johri
Jan 14 at 8:33












4 Answers
4






active

oldest

votes


















2












$begingroup$

I would use logarithmic differentiation. Setting $u(x)=x^2cos x$, we have
begin{align}
frac{f'(x)}{f(x)}&=frac{u'(x)}{u(x)ln u(x)}-frac12frac{-2x}{1-x^2} =frac{2xcos x-x^2sin x}{x^2cos x,ln(x^2cos x)}-frac12frac{-2x}{1-x^2}\[1ex]
&=frac{2cos x-xsin x}{xcos x,ln(x^2cos x)}+frac x{1-x^2},
end{align}
whence
$$f'(x)=frac{f'(x)}{f(x)},f(x)=frac{2cos x-xsin x}{xcos xsqrt{1-x^2}}+frac{xln(x^2cos x)}{(1-x^2)^{3/2}}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I mean thank you so much for the different approch $ dfrac{xlnleft(x^2cosleft(xright)right)}{left(1-x^2right)^frac{3}{2}}+dfrac{2xcosleft(xright)-x^2sinleft(xright)}{x^2sqrt{1-x^2}cosleft(xright)} $
    $endgroup$
    – Tariro Manyika
    Jan 14 at 9:05












  • $begingroup$
    @TariroManyika: I don't agree with the $x^2$ in the denominator: there was a simpilfication by $x$ in the computation line above.
    $endgroup$
    – Bernard
    Jan 14 at 9:10










  • $begingroup$
    Yeah I realized my error , you are 100% correct
    $endgroup$
    – Tariro Manyika
    Jan 14 at 9:12










  • $begingroup$
    Your method is actually much fast than what I ended up having to do , I will keep this cool logarithmic trick for a rainy day
    $endgroup$
    – Tariro Manyika
    Jan 14 at 9:13



















1












$begingroup$


I thought about the quotient rule




That's right!



There's a quotient of the functions $g(x)= ln left( x ^ { 2 } cos ( x ) right)$ and $h(x)= sqrt { 1 - x ^ { 2 } }$ so if you can find $g'(x)$ and $h'(x)$, the derivative $f'(x)$ follows as:
$$f'(x)=frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$$
I don't know why you think you'd need another quotient rule afterwards?



For $g'(x)$ and $h'(x)$ you need to apply the chain rule and for the "inner part" $x ^ { 2 } cos ( x )$ you'll need the product rule as well: break it down step by step.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $dfrac{frac{xlnleft(x^2cosleft(xright)right)}{sqrt{1-x^2}}+frac{sqrt{1-x^2}left(2xcosleft(xright)-x^2sinleft(xright)right)}{x^2cosleft(xright)}}{1-x^2}$
    $endgroup$
    – Tariro Manyika
    Jan 14 at 9:07



















1












$begingroup$

The quotient rule is a good idea. First differentiate $ln(x^2 cos (x))$ with the chain rule and the product rule.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    We have $${dover dx}ln {x^2cos x}={1over x^2cos x}cdot {(2xcos x-x^2sin x)}\{dover dx}sqrt{1-x^2}=-{xover sqrt{1-x^2}}$$therefore by defining $g(x)=ln x^2cos x$ and $h(x)=sqrt{1-x^2}$ and using $left({gover h}right)'={g'h-gh'over h^2}$ we finally obtain$$f'(x){=left({gover h}right)'=left({g'h-gh'over h^2}right)(x)\={left({1over x^2cos x}cdot {(2xcos x-x^2sin x)}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2over x} {-tan x}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2} {-xtan x}right)(1-x^2)+x^2ln x^2cos xover x(1-x^2)sqrt{1-x^2}}}$$ for $|x|<1$ and $xne 0$.






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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      I would use logarithmic differentiation. Setting $u(x)=x^2cos x$, we have
      begin{align}
      frac{f'(x)}{f(x)}&=frac{u'(x)}{u(x)ln u(x)}-frac12frac{-2x}{1-x^2} =frac{2xcos x-x^2sin x}{x^2cos x,ln(x^2cos x)}-frac12frac{-2x}{1-x^2}\[1ex]
      &=frac{2cos x-xsin x}{xcos x,ln(x^2cos x)}+frac x{1-x^2},
      end{align}
      whence
      $$f'(x)=frac{f'(x)}{f(x)},f(x)=frac{2cos x-xsin x}{xcos xsqrt{1-x^2}}+frac{xln(x^2cos x)}{(1-x^2)^{3/2}}.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I mean thank you so much for the different approch $ dfrac{xlnleft(x^2cosleft(xright)right)}{left(1-x^2right)^frac{3}{2}}+dfrac{2xcosleft(xright)-x^2sinleft(xright)}{x^2sqrt{1-x^2}cosleft(xright)} $
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:05












      • $begingroup$
        @TariroManyika: I don't agree with the $x^2$ in the denominator: there was a simpilfication by $x$ in the computation line above.
        $endgroup$
        – Bernard
        Jan 14 at 9:10










      • $begingroup$
        Yeah I realized my error , you are 100% correct
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:12










      • $begingroup$
        Your method is actually much fast than what I ended up having to do , I will keep this cool logarithmic trick for a rainy day
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:13
















      2












      $begingroup$

      I would use logarithmic differentiation. Setting $u(x)=x^2cos x$, we have
      begin{align}
      frac{f'(x)}{f(x)}&=frac{u'(x)}{u(x)ln u(x)}-frac12frac{-2x}{1-x^2} =frac{2xcos x-x^2sin x}{x^2cos x,ln(x^2cos x)}-frac12frac{-2x}{1-x^2}\[1ex]
      &=frac{2cos x-xsin x}{xcos x,ln(x^2cos x)}+frac x{1-x^2},
      end{align}
      whence
      $$f'(x)=frac{f'(x)}{f(x)},f(x)=frac{2cos x-xsin x}{xcos xsqrt{1-x^2}}+frac{xln(x^2cos x)}{(1-x^2)^{3/2}}.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I mean thank you so much for the different approch $ dfrac{xlnleft(x^2cosleft(xright)right)}{left(1-x^2right)^frac{3}{2}}+dfrac{2xcosleft(xright)-x^2sinleft(xright)}{x^2sqrt{1-x^2}cosleft(xright)} $
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:05












      • $begingroup$
        @TariroManyika: I don't agree with the $x^2$ in the denominator: there was a simpilfication by $x$ in the computation line above.
        $endgroup$
        – Bernard
        Jan 14 at 9:10










      • $begingroup$
        Yeah I realized my error , you are 100% correct
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:12










      • $begingroup$
        Your method is actually much fast than what I ended up having to do , I will keep this cool logarithmic trick for a rainy day
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:13














      2












      2








      2





      $begingroup$

      I would use logarithmic differentiation. Setting $u(x)=x^2cos x$, we have
      begin{align}
      frac{f'(x)}{f(x)}&=frac{u'(x)}{u(x)ln u(x)}-frac12frac{-2x}{1-x^2} =frac{2xcos x-x^2sin x}{x^2cos x,ln(x^2cos x)}-frac12frac{-2x}{1-x^2}\[1ex]
      &=frac{2cos x-xsin x}{xcos x,ln(x^2cos x)}+frac x{1-x^2},
      end{align}
      whence
      $$f'(x)=frac{f'(x)}{f(x)},f(x)=frac{2cos x-xsin x}{xcos xsqrt{1-x^2}}+frac{xln(x^2cos x)}{(1-x^2)^{3/2}}.$$






      share|cite|improve this answer









      $endgroup$



      I would use logarithmic differentiation. Setting $u(x)=x^2cos x$, we have
      begin{align}
      frac{f'(x)}{f(x)}&=frac{u'(x)}{u(x)ln u(x)}-frac12frac{-2x}{1-x^2} =frac{2xcos x-x^2sin x}{x^2cos x,ln(x^2cos x)}-frac12frac{-2x}{1-x^2}\[1ex]
      &=frac{2cos x-xsin x}{xcos x,ln(x^2cos x)}+frac x{1-x^2},
      end{align}
      whence
      $$f'(x)=frac{f'(x)}{f(x)},f(x)=frac{2cos x-xsin x}{xcos xsqrt{1-x^2}}+frac{xln(x^2cos x)}{(1-x^2)^{3/2}}.$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 14 at 9:02









      BernardBernard

      123k741117




      123k741117












      • $begingroup$
        I mean thank you so much for the different approch $ dfrac{xlnleft(x^2cosleft(xright)right)}{left(1-x^2right)^frac{3}{2}}+dfrac{2xcosleft(xright)-x^2sinleft(xright)}{x^2sqrt{1-x^2}cosleft(xright)} $
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:05












      • $begingroup$
        @TariroManyika: I don't agree with the $x^2$ in the denominator: there was a simpilfication by $x$ in the computation line above.
        $endgroup$
        – Bernard
        Jan 14 at 9:10










      • $begingroup$
        Yeah I realized my error , you are 100% correct
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:12










      • $begingroup$
        Your method is actually much fast than what I ended up having to do , I will keep this cool logarithmic trick for a rainy day
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:13


















      • $begingroup$
        I mean thank you so much for the different approch $ dfrac{xlnleft(x^2cosleft(xright)right)}{left(1-x^2right)^frac{3}{2}}+dfrac{2xcosleft(xright)-x^2sinleft(xright)}{x^2sqrt{1-x^2}cosleft(xright)} $
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:05












      • $begingroup$
        @TariroManyika: I don't agree with the $x^2$ in the denominator: there was a simpilfication by $x$ in the computation line above.
        $endgroup$
        – Bernard
        Jan 14 at 9:10










      • $begingroup$
        Yeah I realized my error , you are 100% correct
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:12










      • $begingroup$
        Your method is actually much fast than what I ended up having to do , I will keep this cool logarithmic trick for a rainy day
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:13
















      $begingroup$
      I mean thank you so much for the different approch $ dfrac{xlnleft(x^2cosleft(xright)right)}{left(1-x^2right)^frac{3}{2}}+dfrac{2xcosleft(xright)-x^2sinleft(xright)}{x^2sqrt{1-x^2}cosleft(xright)} $
      $endgroup$
      – Tariro Manyika
      Jan 14 at 9:05






      $begingroup$
      I mean thank you so much for the different approch $ dfrac{xlnleft(x^2cosleft(xright)right)}{left(1-x^2right)^frac{3}{2}}+dfrac{2xcosleft(xright)-x^2sinleft(xright)}{x^2sqrt{1-x^2}cosleft(xright)} $
      $endgroup$
      – Tariro Manyika
      Jan 14 at 9:05














      $begingroup$
      @TariroManyika: I don't agree with the $x^2$ in the denominator: there was a simpilfication by $x$ in the computation line above.
      $endgroup$
      – Bernard
      Jan 14 at 9:10




      $begingroup$
      @TariroManyika: I don't agree with the $x^2$ in the denominator: there was a simpilfication by $x$ in the computation line above.
      $endgroup$
      – Bernard
      Jan 14 at 9:10












      $begingroup$
      Yeah I realized my error , you are 100% correct
      $endgroup$
      – Tariro Manyika
      Jan 14 at 9:12




      $begingroup$
      Yeah I realized my error , you are 100% correct
      $endgroup$
      – Tariro Manyika
      Jan 14 at 9:12












      $begingroup$
      Your method is actually much fast than what I ended up having to do , I will keep this cool logarithmic trick for a rainy day
      $endgroup$
      – Tariro Manyika
      Jan 14 at 9:13




      $begingroup$
      Your method is actually much fast than what I ended up having to do , I will keep this cool logarithmic trick for a rainy day
      $endgroup$
      – Tariro Manyika
      Jan 14 at 9:13











      1












      $begingroup$


      I thought about the quotient rule




      That's right!



      There's a quotient of the functions $g(x)= ln left( x ^ { 2 } cos ( x ) right)$ and $h(x)= sqrt { 1 - x ^ { 2 } }$ so if you can find $g'(x)$ and $h'(x)$, the derivative $f'(x)$ follows as:
      $$f'(x)=frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$$
      I don't know why you think you'd need another quotient rule afterwards?



      For $g'(x)$ and $h'(x)$ you need to apply the chain rule and for the "inner part" $x ^ { 2 } cos ( x )$ you'll need the product rule as well: break it down step by step.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        $dfrac{frac{xlnleft(x^2cosleft(xright)right)}{sqrt{1-x^2}}+frac{sqrt{1-x^2}left(2xcosleft(xright)-x^2sinleft(xright)right)}{x^2cosleft(xright)}}{1-x^2}$
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:07
















      1












      $begingroup$


      I thought about the quotient rule




      That's right!



      There's a quotient of the functions $g(x)= ln left( x ^ { 2 } cos ( x ) right)$ and $h(x)= sqrt { 1 - x ^ { 2 } }$ so if you can find $g'(x)$ and $h'(x)$, the derivative $f'(x)$ follows as:
      $$f'(x)=frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$$
      I don't know why you think you'd need another quotient rule afterwards?



      For $g'(x)$ and $h'(x)$ you need to apply the chain rule and for the "inner part" $x ^ { 2 } cos ( x )$ you'll need the product rule as well: break it down step by step.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        $dfrac{frac{xlnleft(x^2cosleft(xright)right)}{sqrt{1-x^2}}+frac{sqrt{1-x^2}left(2xcosleft(xright)-x^2sinleft(xright)right)}{x^2cosleft(xright)}}{1-x^2}$
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:07














      1












      1








      1





      $begingroup$


      I thought about the quotient rule




      That's right!



      There's a quotient of the functions $g(x)= ln left( x ^ { 2 } cos ( x ) right)$ and $h(x)= sqrt { 1 - x ^ { 2 } }$ so if you can find $g'(x)$ and $h'(x)$, the derivative $f'(x)$ follows as:
      $$f'(x)=frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$$
      I don't know why you think you'd need another quotient rule afterwards?



      For $g'(x)$ and $h'(x)$ you need to apply the chain rule and for the "inner part" $x ^ { 2 } cos ( x )$ you'll need the product rule as well: break it down step by step.






      share|cite|improve this answer









      $endgroup$




      I thought about the quotient rule




      That's right!



      There's a quotient of the functions $g(x)= ln left( x ^ { 2 } cos ( x ) right)$ and $h(x)= sqrt { 1 - x ^ { 2 } }$ so if you can find $g'(x)$ and $h'(x)$, the derivative $f'(x)$ follows as:
      $$f'(x)=frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$$
      I don't know why you think you'd need another quotient rule afterwards?



      For $g'(x)$ and $h'(x)$ you need to apply the chain rule and for the "inner part" $x ^ { 2 } cos ( x )$ you'll need the product rule as well: break it down step by step.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 14 at 8:34









      StackTDStackTD

      23.9k2254




      23.9k2254












      • $begingroup$
        $dfrac{frac{xlnleft(x^2cosleft(xright)right)}{sqrt{1-x^2}}+frac{sqrt{1-x^2}left(2xcosleft(xright)-x^2sinleft(xright)right)}{x^2cosleft(xright)}}{1-x^2}$
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:07


















      • $begingroup$
        $dfrac{frac{xlnleft(x^2cosleft(xright)right)}{sqrt{1-x^2}}+frac{sqrt{1-x^2}left(2xcosleft(xright)-x^2sinleft(xright)right)}{x^2cosleft(xright)}}{1-x^2}$
        $endgroup$
        – Tariro Manyika
        Jan 14 at 9:07
















      $begingroup$
      $dfrac{frac{xlnleft(x^2cosleft(xright)right)}{sqrt{1-x^2}}+frac{sqrt{1-x^2}left(2xcosleft(xright)-x^2sinleft(xright)right)}{x^2cosleft(xright)}}{1-x^2}$
      $endgroup$
      – Tariro Manyika
      Jan 14 at 9:07




      $begingroup$
      $dfrac{frac{xlnleft(x^2cosleft(xright)right)}{sqrt{1-x^2}}+frac{sqrt{1-x^2}left(2xcosleft(xright)-x^2sinleft(xright)right)}{x^2cosleft(xright)}}{1-x^2}$
      $endgroup$
      – Tariro Manyika
      Jan 14 at 9:07











      1












      $begingroup$

      The quotient rule is a good idea. First differentiate $ln(x^2 cos (x))$ with the chain rule and the product rule.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        The quotient rule is a good idea. First differentiate $ln(x^2 cos (x))$ with the chain rule and the product rule.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          The quotient rule is a good idea. First differentiate $ln(x^2 cos (x))$ with the chain rule and the product rule.






          share|cite|improve this answer











          $endgroup$



          The quotient rule is a good idea. First differentiate $ln(x^2 cos (x))$ with the chain rule and the product rule.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 8:45









          Bernard

          123k741117




          123k741117










          answered Jan 14 at 8:33









          FredFred

          48.6k11849




          48.6k11849























              1












              $begingroup$

              We have $${dover dx}ln {x^2cos x}={1over x^2cos x}cdot {(2xcos x-x^2sin x)}\{dover dx}sqrt{1-x^2}=-{xover sqrt{1-x^2}}$$therefore by defining $g(x)=ln x^2cos x$ and $h(x)=sqrt{1-x^2}$ and using $left({gover h}right)'={g'h-gh'over h^2}$ we finally obtain$$f'(x){=left({gover h}right)'=left({g'h-gh'over h^2}right)(x)\={left({1over x^2cos x}cdot {(2xcos x-x^2sin x)}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2over x} {-tan x}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2} {-xtan x}right)(1-x^2)+x^2ln x^2cos xover x(1-x^2)sqrt{1-x^2}}}$$ for $|x|<1$ and $xne 0$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                We have $${dover dx}ln {x^2cos x}={1over x^2cos x}cdot {(2xcos x-x^2sin x)}\{dover dx}sqrt{1-x^2}=-{xover sqrt{1-x^2}}$$therefore by defining $g(x)=ln x^2cos x$ and $h(x)=sqrt{1-x^2}$ and using $left({gover h}right)'={g'h-gh'over h^2}$ we finally obtain$$f'(x){=left({gover h}right)'=left({g'h-gh'over h^2}right)(x)\={left({1over x^2cos x}cdot {(2xcos x-x^2sin x)}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2over x} {-tan x}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2} {-xtan x}right)(1-x^2)+x^2ln x^2cos xover x(1-x^2)sqrt{1-x^2}}}$$ for $|x|<1$ and $xne 0$.






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                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  We have $${dover dx}ln {x^2cos x}={1over x^2cos x}cdot {(2xcos x-x^2sin x)}\{dover dx}sqrt{1-x^2}=-{xover sqrt{1-x^2}}$$therefore by defining $g(x)=ln x^2cos x$ and $h(x)=sqrt{1-x^2}$ and using $left({gover h}right)'={g'h-gh'over h^2}$ we finally obtain$$f'(x){=left({gover h}right)'=left({g'h-gh'over h^2}right)(x)\={left({1over x^2cos x}cdot {(2xcos x-x^2sin x)}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2over x} {-tan x}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2} {-xtan x}right)(1-x^2)+x^2ln x^2cos xover x(1-x^2)sqrt{1-x^2}}}$$ for $|x|<1$ and $xne 0$.






                  share|cite|improve this answer











                  $endgroup$



                  We have $${dover dx}ln {x^2cos x}={1over x^2cos x}cdot {(2xcos x-x^2sin x)}\{dover dx}sqrt{1-x^2}=-{xover sqrt{1-x^2}}$$therefore by defining $g(x)=ln x^2cos x$ and $h(x)=sqrt{1-x^2}$ and using $left({gover h}right)'={g'h-gh'over h^2}$ we finally obtain$$f'(x){=left({gover h}right)'=left({g'h-gh'over h^2}right)(x)\={left({1over x^2cos x}cdot {(2xcos x-x^2sin x)}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2over x} {-tan x}right)sqrt{1-x^2}+left({xover sqrt{1-x^2}}right)ln x^2cos xover 1-x^2}\={left({2} {-xtan x}right)(1-x^2)+x^2ln x^2cos xover x(1-x^2)sqrt{1-x^2}}}$$ for $|x|<1$ and $xne 0$.







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                  share|cite|improve this answer








                  edited Jan 14 at 9:22

























                  answered Jan 14 at 9:15









                  Mostafa AyazMostafa Ayaz

                  17.2k31039




                  17.2k31039






























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