Trouble taking derivative












0














$$p(x) = frac{RTx}{1-bx} -ax^2$$



If $R$ and $T$ are constants, how can I find the derivative of this?



What I obtained:



$p'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}bRTx - 2ax$



Is this correct?



Thanks!






derivatives







share|cite|improve this question




















  • 1




    If you show your steps someone could advise you better. To me it doesn't look like you used the quotient rule on first term.
    – coffeemath
    Nov 27 at 1:03








  • 1




    Almost correct, but not quite. In executing the product rule on the first term, note that the derivative of $(1-xb)^{-1}$ is $$(-1)(-b)(1-xb)^{-2} = b(1-xb)^{-2}$$ per the chain rule. So you wouldn't have that $RTx$ in the second term of your expression - just $RT$ - and the sign of the term would be positive (not negative).
    – Eevee Trainer
    Nov 27 at 1:03
















0














$$p(x) = frac{RTx}{1-bx} -ax^2$$



If $R$ and $T$ are constants, how can I find the derivative of this?



What I obtained:



$p'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}bRTx - 2ax$



Is this correct?



Thanks!






derivatives







share|cite|improve this question




















  • 1




    If you show your steps someone could advise you better. To me it doesn't look like you used the quotient rule on first term.
    – coffeemath
    Nov 27 at 1:03








  • 1




    Almost correct, but not quite. In executing the product rule on the first term, note that the derivative of $(1-xb)^{-1}$ is $$(-1)(-b)(1-xb)^{-2} = b(1-xb)^{-2}$$ per the chain rule. So you wouldn't have that $RTx$ in the second term of your expression - just $RT$ - and the sign of the term would be positive (not negative).
    – Eevee Trainer
    Nov 27 at 1:03














0












0








0


1





$$p(x) = frac{RTx}{1-bx} -ax^2$$



If $R$ and $T$ are constants, how can I find the derivative of this?



What I obtained:



$p'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}bRTx - 2ax$



Is this correct?



Thanks!






derivatives







share|cite|improve this question















$$p(x) = frac{RTx}{1-bx} -ax^2$$



If $R$ and $T$ are constants, how can I find the derivative of this?



What I obtained:



$p'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}bRTx - 2ax$



Is this correct?



Thanks!






derivatives




derivatives






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 1:32









Eevee Trainer

4,090530




4,090530










asked Nov 27 at 0:57









M Do

124




124








  • 1




    If you show your steps someone could advise you better. To me it doesn't look like you used the quotient rule on first term.
    – coffeemath
    Nov 27 at 1:03








  • 1




    Almost correct, but not quite. In executing the product rule on the first term, note that the derivative of $(1-xb)^{-1}$ is $$(-1)(-b)(1-xb)^{-2} = b(1-xb)^{-2}$$ per the chain rule. So you wouldn't have that $RTx$ in the second term of your expression - just $RT$ - and the sign of the term would be positive (not negative).
    – Eevee Trainer
    Nov 27 at 1:03














  • 1




    If you show your steps someone could advise you better. To me it doesn't look like you used the quotient rule on first term.
    – coffeemath
    Nov 27 at 1:03








  • 1




    Almost correct, but not quite. In executing the product rule on the first term, note that the derivative of $(1-xb)^{-1}$ is $$(-1)(-b)(1-xb)^{-2} = b(1-xb)^{-2}$$ per the chain rule. So you wouldn't have that $RTx$ in the second term of your expression - just $RT$ - and the sign of the term would be positive (not negative).
    – Eevee Trainer
    Nov 27 at 1:03








1




1




If you show your steps someone could advise you better. To me it doesn't look like you used the quotient rule on first term.
– coffeemath
Nov 27 at 1:03






If you show your steps someone could advise you better. To me it doesn't look like you used the quotient rule on first term.
– coffeemath
Nov 27 at 1:03






1




1




Almost correct, but not quite. In executing the product rule on the first term, note that the derivative of $(1-xb)^{-1}$ is $$(-1)(-b)(1-xb)^{-2} = b(1-xb)^{-2}$$ per the chain rule. So you wouldn't have that $RTx$ in the second term of your expression - just $RT$ - and the sign of the term would be positive (not negative).
– Eevee Trainer
Nov 27 at 1:03




Almost correct, but not quite. In executing the product rule on the first term, note that the derivative of $(1-xb)^{-1}$ is $$(-1)(-b)(1-xb)^{-2} = b(1-xb)^{-2}$$ per the chain rule. So you wouldn't have that $RTx$ in the second term of your expression - just $RT$ - and the sign of the term would be positive (not negative).
– Eevee Trainer
Nov 27 at 1:03










3 Answers
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0














$$p(x) = frac{RTx}{1-bx} -ax^2=(RTx)(1-bx)^{-1}-ax^2$$



begin{align}p'(x)&=(RT)(1-bx)^{-1}+(RTx)(-1)(1-bx)^{-2}(-b)-2ax \
&=(RT)(1-bx)^{-1}+(bRTx)(1-bx)^{-2}-2axend{align}






share|cite|improve this answer





























    0














    I would say you should simplify a little bit more.



    $f'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}(-b)(RTx)- 2ax\
    RT(1-xb)^{-2}(1-xb + xb) - 2ax\
    frac {RT}{(1-xb)^2}-2ax$






    share|cite|improve this answer





























      0














      Use the quotient rule: ($frac {f(x)}{g(x)} = frac {f'(x)g(x) - f(x)g'(x)}{g^2(x)}$)



      the sum rule: ($(f(x) + g(x))' = f'(x) + g'(x)$.)



      constant rule: ($(af(x))' = af'(x)$)



      And power rule: ($[x^k]' = kx^{k-1}$)



      And the entire thing is simple:



      $p(x) = frac{RTx}{1-bx} -ax^2$ so



      $p'(x) = frac{[RTx]'(1-bx) -RTx[1-bx]'}{(1-bx)^2} -[ax^2]'=$



      $ frac {RT(1-bx)-RTx(-b)}{(1-bx)^2} - 2ax=$



      $frac {RT - RTbx + RTxb}{(1-bx)^2} -2ax = frac {RT}{(1-bx)^2} - 2ax$.






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
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        0














        $$p(x) = frac{RTx}{1-bx} -ax^2=(RTx)(1-bx)^{-1}-ax^2$$



        begin{align}p'(x)&=(RT)(1-bx)^{-1}+(RTx)(-1)(1-bx)^{-2}(-b)-2ax \
        &=(RT)(1-bx)^{-1}+(bRTx)(1-bx)^{-2}-2axend{align}






        share|cite|improve this answer


























          0














          $$p(x) = frac{RTx}{1-bx} -ax^2=(RTx)(1-bx)^{-1}-ax^2$$



          begin{align}p'(x)&=(RT)(1-bx)^{-1}+(RTx)(-1)(1-bx)^{-2}(-b)-2ax \
          &=(RT)(1-bx)^{-1}+(bRTx)(1-bx)^{-2}-2axend{align}






          share|cite|improve this answer
























            0












            0








            0






            $$p(x) = frac{RTx}{1-bx} -ax^2=(RTx)(1-bx)^{-1}-ax^2$$



            begin{align}p'(x)&=(RT)(1-bx)^{-1}+(RTx)(-1)(1-bx)^{-2}(-b)-2ax \
            &=(RT)(1-bx)^{-1}+(bRTx)(1-bx)^{-2}-2axend{align}






            share|cite|improve this answer












            $$p(x) = frac{RTx}{1-bx} -ax^2=(RTx)(1-bx)^{-1}-ax^2$$



            begin{align}p'(x)&=(RT)(1-bx)^{-1}+(RTx)(-1)(1-bx)^{-2}(-b)-2ax \
            &=(RT)(1-bx)^{-1}+(bRTx)(1-bx)^{-2}-2axend{align}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 27 at 1:09









            Siong Thye Goh

            99k1464116




            99k1464116























                0














                I would say you should simplify a little bit more.



                $f'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}(-b)(RTx)- 2ax\
                RT(1-xb)^{-2}(1-xb + xb) - 2ax\
                frac {RT}{(1-xb)^2}-2ax$






                share|cite|improve this answer


























                  0














                  I would say you should simplify a little bit more.



                  $f'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}(-b)(RTx)- 2ax\
                  RT(1-xb)^{-2}(1-xb + xb) - 2ax\
                  frac {RT}{(1-xb)^2}-2ax$






                  share|cite|improve this answer
























                    0












                    0








                    0






                    I would say you should simplify a little bit more.



                    $f'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}(-b)(RTx)- 2ax\
                    RT(1-xb)^{-2}(1-xb + xb) - 2ax\
                    frac {RT}{(1-xb)^2}-2ax$






                    share|cite|improve this answer












                    I would say you should simplify a little bit more.



                    $f'(x) = RT(1-xb)^{-1} + (-1)(1-xb)^{-2}(-b)(RTx)- 2ax\
                    RT(1-xb)^{-2}(1-xb + xb) - 2ax\
                    frac {RT}{(1-xb)^2}-2ax$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 27 at 2:05









                    Doug M

                    43.9k31854




                    43.9k31854























                        0














                        Use the quotient rule: ($frac {f(x)}{g(x)} = frac {f'(x)g(x) - f(x)g'(x)}{g^2(x)}$)



                        the sum rule: ($(f(x) + g(x))' = f'(x) + g'(x)$.)



                        constant rule: ($(af(x))' = af'(x)$)



                        And power rule: ($[x^k]' = kx^{k-1}$)



                        And the entire thing is simple:



                        $p(x) = frac{RTx}{1-bx} -ax^2$ so



                        $p'(x) = frac{[RTx]'(1-bx) -RTx[1-bx]'}{(1-bx)^2} -[ax^2]'=$



                        $ frac {RT(1-bx)-RTx(-b)}{(1-bx)^2} - 2ax=$



                        $frac {RT - RTbx + RTxb}{(1-bx)^2} -2ax = frac {RT}{(1-bx)^2} - 2ax$.






                        share|cite|improve this answer


























                          0














                          Use the quotient rule: ($frac {f(x)}{g(x)} = frac {f'(x)g(x) - f(x)g'(x)}{g^2(x)}$)



                          the sum rule: ($(f(x) + g(x))' = f'(x) + g'(x)$.)



                          constant rule: ($(af(x))' = af'(x)$)



                          And power rule: ($[x^k]' = kx^{k-1}$)



                          And the entire thing is simple:



                          $p(x) = frac{RTx}{1-bx} -ax^2$ so



                          $p'(x) = frac{[RTx]'(1-bx) -RTx[1-bx]'}{(1-bx)^2} -[ax^2]'=$



                          $ frac {RT(1-bx)-RTx(-b)}{(1-bx)^2} - 2ax=$



                          $frac {RT - RTbx + RTxb}{(1-bx)^2} -2ax = frac {RT}{(1-bx)^2} - 2ax$.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            Use the quotient rule: ($frac {f(x)}{g(x)} = frac {f'(x)g(x) - f(x)g'(x)}{g^2(x)}$)



                            the sum rule: ($(f(x) + g(x))' = f'(x) + g'(x)$.)



                            constant rule: ($(af(x))' = af'(x)$)



                            And power rule: ($[x^k]' = kx^{k-1}$)



                            And the entire thing is simple:



                            $p(x) = frac{RTx}{1-bx} -ax^2$ so



                            $p'(x) = frac{[RTx]'(1-bx) -RTx[1-bx]'}{(1-bx)^2} -[ax^2]'=$



                            $ frac {RT(1-bx)-RTx(-b)}{(1-bx)^2} - 2ax=$



                            $frac {RT - RTbx + RTxb}{(1-bx)^2} -2ax = frac {RT}{(1-bx)^2} - 2ax$.






                            share|cite|improve this answer












                            Use the quotient rule: ($frac {f(x)}{g(x)} = frac {f'(x)g(x) - f(x)g'(x)}{g^2(x)}$)



                            the sum rule: ($(f(x) + g(x))' = f'(x) + g'(x)$.)



                            constant rule: ($(af(x))' = af'(x)$)



                            And power rule: ($[x^k]' = kx^{k-1}$)



                            And the entire thing is simple:



                            $p(x) = frac{RTx}{1-bx} -ax^2$ so



                            $p'(x) = frac{[RTx]'(1-bx) -RTx[1-bx]'}{(1-bx)^2} -[ax^2]'=$



                            $ frac {RT(1-bx)-RTx(-b)}{(1-bx)^2} - 2ax=$



                            $frac {RT - RTbx + RTxb}{(1-bx)^2} -2ax = frac {RT}{(1-bx)^2} - 2ax$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 27 at 3:06









                            fleablood

                            68.1k22684




                            68.1k22684






























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