Let $y=(f(u)+3x)^2$ and $u=x^3-2x$. If $f(4)=6$ and $frac{dy}{dx}=18$ when $x=2$, find $f'(4)$












2












$begingroup$



Let $y=(f(u)+3x)^2$ and $u=x^3-2x$. If $f(4)=6$ and $frac{dy}{dx}=18$ when $x=2$, find $f'(4)$.




I'm sorry if I'm asking this beginner question, but I'm really confused about how to solve this problem. I've tried to substitute $u=4$ and other things that I thought might work. I'm kinda stuck.



I've got $f'(4)=-frac{27}6$ but in the textbook it is $-frac9{40}$.






derivatives







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$endgroup$








  • 1




    $begingroup$
    How did you get the answer? It might be a simple math mistake
    $endgroup$
    – Andrei
    Jan 13 at 2:22






  • 1




    $begingroup$
    And why do you have twice $f(4)=6$ in the title?
    $endgroup$
    – Andrei
    Jan 13 at 2:24










  • $begingroup$
    @Andrei where ?
    $endgroup$
    – Zein IF
    Jan 13 at 6:48
















2












$begingroup$



Let $y=(f(u)+3x)^2$ and $u=x^3-2x$. If $f(4)=6$ and $frac{dy}{dx}=18$ when $x=2$, find $f'(4)$.




I'm sorry if I'm asking this beginner question, but I'm really confused about how to solve this problem. I've tried to substitute $u=4$ and other things that I thought might work. I'm kinda stuck.



I've got $f'(4)=-frac{27}6$ but in the textbook it is $-frac9{40}$.






derivatives







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How did you get the answer? It might be a simple math mistake
    $endgroup$
    – Andrei
    Jan 13 at 2:22






  • 1




    $begingroup$
    And why do you have twice $f(4)=6$ in the title?
    $endgroup$
    – Andrei
    Jan 13 at 2:24










  • $begingroup$
    @Andrei where ?
    $endgroup$
    – Zein IF
    Jan 13 at 6:48














2












2








2





$begingroup$



Let $y=(f(u)+3x)^2$ and $u=x^3-2x$. If $f(4)=6$ and $frac{dy}{dx}=18$ when $x=2$, find $f'(4)$.




I'm sorry if I'm asking this beginner question, but I'm really confused about how to solve this problem. I've tried to substitute $u=4$ and other things that I thought might work. I'm kinda stuck.



I've got $f'(4)=-frac{27}6$ but in the textbook it is $-frac9{40}$.






derivatives







share|cite|improve this question











$endgroup$





Let $y=(f(u)+3x)^2$ and $u=x^3-2x$. If $f(4)=6$ and $frac{dy}{dx}=18$ when $x=2$, find $f'(4)$.




I'm sorry if I'm asking this beginner question, but I'm really confused about how to solve this problem. I've tried to substitute $u=4$ and other things that I thought might work. I'm kinda stuck.



I've got $f'(4)=-frac{27}6$ but in the textbook it is $-frac9{40}$.






derivatives




derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 2:46









Parcly Taxel

41.8k1372101




41.8k1372101










asked Jan 13 at 2:04









Zein IFZein IF

134




134








  • 1




    $begingroup$
    How did you get the answer? It might be a simple math mistake
    $endgroup$
    – Andrei
    Jan 13 at 2:22






  • 1




    $begingroup$
    And why do you have twice $f(4)=6$ in the title?
    $endgroup$
    – Andrei
    Jan 13 at 2:24










  • $begingroup$
    @Andrei where ?
    $endgroup$
    – Zein IF
    Jan 13 at 6:48














  • 1




    $begingroup$
    How did you get the answer? It might be a simple math mistake
    $endgroup$
    – Andrei
    Jan 13 at 2:22






  • 1




    $begingroup$
    And why do you have twice $f(4)=6$ in the title?
    $endgroup$
    – Andrei
    Jan 13 at 2:24










  • $begingroup$
    @Andrei where ?
    $endgroup$
    – Zein IF
    Jan 13 at 6:48








1




1




$begingroup$
How did you get the answer? It might be a simple math mistake
$endgroup$
– Andrei
Jan 13 at 2:22




$begingroup$
How did you get the answer? It might be a simple math mistake
$endgroup$
– Andrei
Jan 13 at 2:22




1




1




$begingroup$
And why do you have twice $f(4)=6$ in the title?
$endgroup$
– Andrei
Jan 13 at 2:24




$begingroup$
And why do you have twice $f(4)=6$ in the title?
$endgroup$
– Andrei
Jan 13 at 2:24












$begingroup$
@Andrei where ?
$endgroup$
– Zein IF
Jan 13 at 6:48




$begingroup$
@Andrei where ?
$endgroup$
– Zein IF
Jan 13 at 6:48










3 Answers
3






active

oldest

votes


















0












$begingroup$

Note that for $x=2, u=4$
$$begin{align}y&=(f(u(x))+3x)^2\dfrac{mathrm d}{mathrm dx}(y)&=dfrac{mathrm d}{mathrm dx}(f(u(x))+3x)^2\dfrac{mathrm dy}{mathrm dx}&=2(f(u(x))+3x)cdotdfrac{mathrm d}{mathrm dx}(f(u(x))+3x)\&=2(f(u(x))+3x)(f'(u(x))cdot u'(x)+3)\text{Setting }&x=2...\18&=2(f(u(2))+6)(f'(u(2))cdot u'(2)+3)\9&=(f(4)+6)(f'(4)cdot(3cdot2^2-2)+3)\9&=12(10f'(4)+3)\dfrac34&=10f'(4)+3\10f'(4)&=-dfrac94\f'(4)&=-dfrac9{40}end{align}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh jeez i forgot to differentiate the f(u(x)).
    $endgroup$
    – Zein IF
    Jan 13 at 6:52



















0












$begingroup$

Differentiate with respect to $x$:
$$y'=2(f(u)+3x)(f'(u)u'+3)=2(f(x^3-2x)+3x)(f'(x^3-2x)(3x^2-2)+3)$$
Substitute the known values at $x=2$ and simplify:
$$18=2(f(4)+6)(10f'(4)+3)=2cdot12(10f'(4)+3)$$
$$frac{18}{24}=10f'(4)+3$$
$$f'(4)=frac{3/4-3}{10}=-frac9{40}$$
So the textbook is right after all.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You have $u(2)=4$ so $f(u(2))=f(4)=6$. Let's write $frac{dy}{dx}$:
    $$begin{align}frac{dy}{dx}&=2(f(u)+3x)frac d{dx}(f(u)+3x)\
    &=2(f(u)+3x)left(frac{df(u(x))}{dx}+3right)\&=2(f(u)+3x)left(frac{df(u)}{du}frac{du}{dx}+3right)\&=2(f(u)+3x)left(f'(u)(3x^2-2)+3right)end{align}$$
    Now just plug in $frac{dy}{dx}=18$, $u=4$, $f(4)=6$, and $x=2$.






    share|cite|improve this answer









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      3 Answers
      3






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      3 Answers
      3






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      active

      oldest

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      active

      oldest

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      0












      $begingroup$

      Note that for $x=2, u=4$
      $$begin{align}y&=(f(u(x))+3x)^2\dfrac{mathrm d}{mathrm dx}(y)&=dfrac{mathrm d}{mathrm dx}(f(u(x))+3x)^2\dfrac{mathrm dy}{mathrm dx}&=2(f(u(x))+3x)cdotdfrac{mathrm d}{mathrm dx}(f(u(x))+3x)\&=2(f(u(x))+3x)(f'(u(x))cdot u'(x)+3)\text{Setting }&x=2...\18&=2(f(u(2))+6)(f'(u(2))cdot u'(2)+3)\9&=(f(4)+6)(f'(4)cdot(3cdot2^2-2)+3)\9&=12(10f'(4)+3)\dfrac34&=10f'(4)+3\10f'(4)&=-dfrac94\f'(4)&=-dfrac9{40}end{align}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        oh jeez i forgot to differentiate the f(u(x)).
        $endgroup$
        – Zein IF
        Jan 13 at 6:52
















      0












      $begingroup$

      Note that for $x=2, u=4$
      $$begin{align}y&=(f(u(x))+3x)^2\dfrac{mathrm d}{mathrm dx}(y)&=dfrac{mathrm d}{mathrm dx}(f(u(x))+3x)^2\dfrac{mathrm dy}{mathrm dx}&=2(f(u(x))+3x)cdotdfrac{mathrm d}{mathrm dx}(f(u(x))+3x)\&=2(f(u(x))+3x)(f'(u(x))cdot u'(x)+3)\text{Setting }&x=2...\18&=2(f(u(2))+6)(f'(u(2))cdot u'(2)+3)\9&=(f(4)+6)(f'(4)cdot(3cdot2^2-2)+3)\9&=12(10f'(4)+3)\dfrac34&=10f'(4)+3\10f'(4)&=-dfrac94\f'(4)&=-dfrac9{40}end{align}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        oh jeez i forgot to differentiate the f(u(x)).
        $endgroup$
        – Zein IF
        Jan 13 at 6:52














      0












      0








      0





      $begingroup$

      Note that for $x=2, u=4$
      $$begin{align}y&=(f(u(x))+3x)^2\dfrac{mathrm d}{mathrm dx}(y)&=dfrac{mathrm d}{mathrm dx}(f(u(x))+3x)^2\dfrac{mathrm dy}{mathrm dx}&=2(f(u(x))+3x)cdotdfrac{mathrm d}{mathrm dx}(f(u(x))+3x)\&=2(f(u(x))+3x)(f'(u(x))cdot u'(x)+3)\text{Setting }&x=2...\18&=2(f(u(2))+6)(f'(u(2))cdot u'(2)+3)\9&=(f(4)+6)(f'(4)cdot(3cdot2^2-2)+3)\9&=12(10f'(4)+3)\dfrac34&=10f'(4)+3\10f'(4)&=-dfrac94\f'(4)&=-dfrac9{40}end{align}$$






      share|cite|improve this answer









      $endgroup$



      Note that for $x=2, u=4$
      $$begin{align}y&=(f(u(x))+3x)^2\dfrac{mathrm d}{mathrm dx}(y)&=dfrac{mathrm d}{mathrm dx}(f(u(x))+3x)^2\dfrac{mathrm dy}{mathrm dx}&=2(f(u(x))+3x)cdotdfrac{mathrm d}{mathrm dx}(f(u(x))+3x)\&=2(f(u(x))+3x)(f'(u(x))cdot u'(x)+3)\text{Setting }&x=2...\18&=2(f(u(2))+6)(f'(u(2))cdot u'(2)+3)\9&=(f(4)+6)(f'(4)cdot(3cdot2^2-2)+3)\9&=12(10f'(4)+3)\dfrac34&=10f'(4)+3\10f'(4)&=-dfrac94\f'(4)&=-dfrac9{40}end{align}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 13 at 3:19









      Awnon BhowmikAwnon Bhowmik

      1,22028




      1,22028












      • $begingroup$
        oh jeez i forgot to differentiate the f(u(x)).
        $endgroup$
        – Zein IF
        Jan 13 at 6:52


















      • $begingroup$
        oh jeez i forgot to differentiate the f(u(x)).
        $endgroup$
        – Zein IF
        Jan 13 at 6:52
















      $begingroup$
      oh jeez i forgot to differentiate the f(u(x)).
      $endgroup$
      – Zein IF
      Jan 13 at 6:52




      $begingroup$
      oh jeez i forgot to differentiate the f(u(x)).
      $endgroup$
      – Zein IF
      Jan 13 at 6:52











      0












      $begingroup$

      Differentiate with respect to $x$:
      $$y'=2(f(u)+3x)(f'(u)u'+3)=2(f(x^3-2x)+3x)(f'(x^3-2x)(3x^2-2)+3)$$
      Substitute the known values at $x=2$ and simplify:
      $$18=2(f(4)+6)(10f'(4)+3)=2cdot12(10f'(4)+3)$$
      $$frac{18}{24}=10f'(4)+3$$
      $$f'(4)=frac{3/4-3}{10}=-frac9{40}$$
      So the textbook is right after all.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Differentiate with respect to $x$:
        $$y'=2(f(u)+3x)(f'(u)u'+3)=2(f(x^3-2x)+3x)(f'(x^3-2x)(3x^2-2)+3)$$
        Substitute the known values at $x=2$ and simplify:
        $$18=2(f(4)+6)(10f'(4)+3)=2cdot12(10f'(4)+3)$$
        $$frac{18}{24}=10f'(4)+3$$
        $$f'(4)=frac{3/4-3}{10}=-frac9{40}$$
        So the textbook is right after all.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Differentiate with respect to $x$:
          $$y'=2(f(u)+3x)(f'(u)u'+3)=2(f(x^3-2x)+3x)(f'(x^3-2x)(3x^2-2)+3)$$
          Substitute the known values at $x=2$ and simplify:
          $$18=2(f(4)+6)(10f'(4)+3)=2cdot12(10f'(4)+3)$$
          $$frac{18}{24}=10f'(4)+3$$
          $$f'(4)=frac{3/4-3}{10}=-frac9{40}$$
          So the textbook is right after all.






          share|cite|improve this answer









          $endgroup$



          Differentiate with respect to $x$:
          $$y'=2(f(u)+3x)(f'(u)u'+3)=2(f(x^3-2x)+3x)(f'(x^3-2x)(3x^2-2)+3)$$
          Substitute the known values at $x=2$ and simplify:
          $$18=2(f(4)+6)(10f'(4)+3)=2cdot12(10f'(4)+3)$$
          $$frac{18}{24}=10f'(4)+3$$
          $$f'(4)=frac{3/4-3}{10}=-frac9{40}$$
          So the textbook is right after all.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 3:09









          Parcly TaxelParcly Taxel

          41.8k1372101




          41.8k1372101























              0












              $begingroup$

              You have $u(2)=4$ so $f(u(2))=f(4)=6$. Let's write $frac{dy}{dx}$:
              $$begin{align}frac{dy}{dx}&=2(f(u)+3x)frac d{dx}(f(u)+3x)\
              &=2(f(u)+3x)left(frac{df(u(x))}{dx}+3right)\&=2(f(u)+3x)left(frac{df(u)}{du}frac{du}{dx}+3right)\&=2(f(u)+3x)left(f'(u)(3x^2-2)+3right)end{align}$$
              Now just plug in $frac{dy}{dx}=18$, $u=4$, $f(4)=6$, and $x=2$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You have $u(2)=4$ so $f(u(2))=f(4)=6$. Let's write $frac{dy}{dx}$:
                $$begin{align}frac{dy}{dx}&=2(f(u)+3x)frac d{dx}(f(u)+3x)\
                &=2(f(u)+3x)left(frac{df(u(x))}{dx}+3right)\&=2(f(u)+3x)left(frac{df(u)}{du}frac{du}{dx}+3right)\&=2(f(u)+3x)left(f'(u)(3x^2-2)+3right)end{align}$$
                Now just plug in $frac{dy}{dx}=18$, $u=4$, $f(4)=6$, and $x=2$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You have $u(2)=4$ so $f(u(2))=f(4)=6$. Let's write $frac{dy}{dx}$:
                  $$begin{align}frac{dy}{dx}&=2(f(u)+3x)frac d{dx}(f(u)+3x)\
                  &=2(f(u)+3x)left(frac{df(u(x))}{dx}+3right)\&=2(f(u)+3x)left(frac{df(u)}{du}frac{du}{dx}+3right)\&=2(f(u)+3x)left(f'(u)(3x^2-2)+3right)end{align}$$
                  Now just plug in $frac{dy}{dx}=18$, $u=4$, $f(4)=6$, and $x=2$.






                  share|cite|improve this answer









                  $endgroup$



                  You have $u(2)=4$ so $f(u(2))=f(4)=6$. Let's write $frac{dy}{dx}$:
                  $$begin{align}frac{dy}{dx}&=2(f(u)+3x)frac d{dx}(f(u)+3x)\
                  &=2(f(u)+3x)left(frac{df(u(x))}{dx}+3right)\&=2(f(u)+3x)left(frac{df(u)}{du}frac{du}{dx}+3right)\&=2(f(u)+3x)left(f'(u)(3x^2-2)+3right)end{align}$$
                  Now just plug in $frac{dy}{dx}=18$, $u=4$, $f(4)=6$, and $x=2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 3:11









                  AndreiAndrei

                  12k21126




                  12k21126






























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