Finding the derivative of $y= (ln x)^2$












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In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:



$y=(ln x)^2$.

First, we take the power rule. This would make it $2(ln x)$. Then you multiply it by the derivative of the inside function right? $ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $frac{2ln x}x$. This doesn’t seem correct so I’m a little confused.



$y=frac{3}{sqrt{2x+1}}$



First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.



I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.






derivatives







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    2












    $begingroup$


    In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:



    $y=(ln x)^2$.

    First, we take the power rule. This would make it $2(ln x)$. Then you multiply it by the derivative of the inside function right? $ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $frac{2ln x}x$. This doesn’t seem correct so I’m a little confused.



    $y=frac{3}{sqrt{2x+1}}$



    First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.



    I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.






    derivatives







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:



      $y=(ln x)^2$.

      First, we take the power rule. This would make it $2(ln x)$. Then you multiply it by the derivative of the inside function right? $ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $frac{2ln x}x$. This doesn’t seem correct so I’m a little confused.



      $y=frac{3}{sqrt{2x+1}}$



      First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.



      I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.






      derivatives







      share|cite|improve this question











      $endgroup$




      In my math class, we are beginning to find derivatives of more complex functions. I’ve been trying questions from my textbook as practice. Here are two of them that I’m trying out:



      $y=(ln x)^2$.

      First, we take the power rule. This would make it $2(ln x)$. Then you multiply it by the derivative of the inside function right? $ln x$’s derivative is $1/x$. So, multiplied by $1/x$. This gives us $frac{2ln x}x$. This doesn’t seem correct so I’m a little confused.



      $y=frac{3}{sqrt{2x+1}}$



      First, we get the denominator to the top. We get $y=3(2x+1)^{-1/2}$. Then, we use the power rule. $y=3(x+1/2)$. Then we multiply by the derivative again. Which is $2$, I believe. This gives us $y=6(2x+1)$. Again, this does not seem correct and I’m confused.



      I feel that my mistakes may be from a mix up of steps but I’m not exactly sure where in my process I went wrong.






      derivatives




      derivatives






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      share|cite|improve this question













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      edited Dec 3 '18 at 0:52







      Ella

















      asked Dec 3 '18 at 0:41









      EllaElla

      33111




      33111






















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          $begingroup$

          First one is correct, for the second one



          begin{eqnarray}
          frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
          &=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
          &=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
          &=& -frac{3}{(2x + 1)^{3/2}}
          end{eqnarray}






          share|cite|improve this answer











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            1 Answer
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            $begingroup$

            First one is correct, for the second one



            begin{eqnarray}
            frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
            &=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
            &=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
            &=& -frac{3}{(2x + 1)^{3/2}}
            end{eqnarray}






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              First one is correct, for the second one



              begin{eqnarray}
              frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
              &=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
              &=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
              &=& -frac{3}{(2x + 1)^{3/2}}
              end{eqnarray}






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                First one is correct, for the second one



                begin{eqnarray}
                frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
                &=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
                &=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
                &=& -frac{3}{(2x + 1)^{3/2}}
                end{eqnarray}






                share|cite|improve this answer











                $endgroup$



                First one is correct, for the second one



                begin{eqnarray}
                frac{{rm d}}{{rm d}x} frac{3}{sqrt{2x + 1}} &=& 3frac{{rm d}}{{rm d}x} (2x + 1)^{-1/2} ~~~~mbox{move constant out} \
                &=& -frac{3}{2}(2x + 1)^{-3/2}frac{{rm d}}{{rm d}x}(2x + 1) ~~~~mbox{power rule + chain rule} \
                &=& -frac{3}{2}(2x + 1)^{-3/2}(2)\
                &=& -frac{3}{(2x + 1)^{3/2}}
                end{eqnarray}







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 3 '18 at 1:40









                sound wave

                1618




                1618










                answered Dec 3 '18 at 1:02









                caveraccaverac

                14.3k21130




                14.3k21130






























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