Using chain rule to find dz/dt leaving answer in terms of t












1












$begingroup$


Can anyone tell me if I have done this correctly



Using the chain rule Find $ frac{dz}{dt} $
if $ z=xy^2, x=e^{-3t} , y=-sin(2t) $



Leaving answer in terms of $t$




$ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{partial x}{partial t} + frac{partial z}{partial y} cdot frac{partial y}{partial t}$



$ frac{dz}{dt} = (y^2 cdot -3e^{-3t})+(2xy cdot -2cos(2t))$



$ frac{dz}{dt} = (sin^2(2t)cdot-3e^{-3t})+(2e^{-3t}cdot-sin(2t) cdot-2cos(2t))$



Many thanks in advance






ordinary-differential-equations chain-rule







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$endgroup$








  • 1




    $begingroup$
    Looks fine. One little critique: When writing out the chain rule for $frac{dz}{dt}$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
    $endgroup$
    – user137731
    Sep 28 '16 at 1:56












  • $begingroup$
    oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
    $endgroup$
    – Buddy
    Sep 28 '16 at 1:59










  • $begingroup$
    Correction - should have read $ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{dx}{dt} + frac{partial z}{partial y} cdot frac{dy}{dt}$
    $endgroup$
    – Buddy
    Sep 28 '16 at 2:00


















1












$begingroup$


Can anyone tell me if I have done this correctly



Using the chain rule Find $ frac{dz}{dt} $
if $ z=xy^2, x=e^{-3t} , y=-sin(2t) $



Leaving answer in terms of $t$




$ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{partial x}{partial t} + frac{partial z}{partial y} cdot frac{partial y}{partial t}$



$ frac{dz}{dt} = (y^2 cdot -3e^{-3t})+(2xy cdot -2cos(2t))$



$ frac{dz}{dt} = (sin^2(2t)cdot-3e^{-3t})+(2e^{-3t}cdot-sin(2t) cdot-2cos(2t))$



Many thanks in advance






ordinary-differential-equations chain-rule







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Looks fine. One little critique: When writing out the chain rule for $frac{dz}{dt}$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
    $endgroup$
    – user137731
    Sep 28 '16 at 1:56












  • $begingroup$
    oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
    $endgroup$
    – Buddy
    Sep 28 '16 at 1:59










  • $begingroup$
    Correction - should have read $ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{dx}{dt} + frac{partial z}{partial y} cdot frac{dy}{dt}$
    $endgroup$
    – Buddy
    Sep 28 '16 at 2:00
















1












1








1





$begingroup$


Can anyone tell me if I have done this correctly



Using the chain rule Find $ frac{dz}{dt} $
if $ z=xy^2, x=e^{-3t} , y=-sin(2t) $



Leaving answer in terms of $t$




$ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{partial x}{partial t} + frac{partial z}{partial y} cdot frac{partial y}{partial t}$



$ frac{dz}{dt} = (y^2 cdot -3e^{-3t})+(2xy cdot -2cos(2t))$



$ frac{dz}{dt} = (sin^2(2t)cdot-3e^{-3t})+(2e^{-3t}cdot-sin(2t) cdot-2cos(2t))$



Many thanks in advance






ordinary-differential-equations chain-rule







share|cite|improve this question









$endgroup$




Can anyone tell me if I have done this correctly



Using the chain rule Find $ frac{dz}{dt} $
if $ z=xy^2, x=e^{-3t} , y=-sin(2t) $



Leaving answer in terms of $t$




$ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{partial x}{partial t} + frac{partial z}{partial y} cdot frac{partial y}{partial t}$



$ frac{dz}{dt} = (y^2 cdot -3e^{-3t})+(2xy cdot -2cos(2t))$



$ frac{dz}{dt} = (sin^2(2t)cdot-3e^{-3t})+(2e^{-3t}cdot-sin(2t) cdot-2cos(2t))$



Many thanks in advance






ordinary-differential-equations chain-rule




ordinary-differential-equations chain-rule






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 28 '16 at 1:52









BuddyBuddy

4328




4328








  • 1




    $begingroup$
    Looks fine. One little critique: When writing out the chain rule for $frac{dz}{dt}$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
    $endgroup$
    – user137731
    Sep 28 '16 at 1:56












  • $begingroup$
    oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
    $endgroup$
    – Buddy
    Sep 28 '16 at 1:59










  • $begingroup$
    Correction - should have read $ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{dx}{dt} + frac{partial z}{partial y} cdot frac{dy}{dt}$
    $endgroup$
    – Buddy
    Sep 28 '16 at 2:00
















  • 1




    $begingroup$
    Looks fine. One little critique: When writing out the chain rule for $frac{dz}{dt}$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
    $endgroup$
    – user137731
    Sep 28 '16 at 1:56












  • $begingroup$
    oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
    $endgroup$
    – Buddy
    Sep 28 '16 at 1:59










  • $begingroup$
    Correction - should have read $ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{dx}{dt} + frac{partial z}{partial y} cdot frac{dy}{dt}$
    $endgroup$
    – Buddy
    Sep 28 '16 at 2:00










1




1




$begingroup$
Looks fine. One little critique: When writing out the chain rule for $frac{dz}{dt}$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
$endgroup$
– user137731
Sep 28 '16 at 1:56






$begingroup$
Looks fine. One little critique: When writing out the chain rule for $frac{dz}{dt}$, you may want to write the derivatives of $x$ and $y$ as total derivatives rather than partial derivatives since each is a function of only $t$.
$endgroup$
– user137731
Sep 28 '16 at 1:56














$begingroup$
oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
$endgroup$
– Buddy
Sep 28 '16 at 1:59




$begingroup$
oh. you are absolutley correct thanks for pointing that out. That was where i was getting confused but was just me not paying attention to detail. You are a champ.
$endgroup$
– Buddy
Sep 28 '16 at 1:59












$begingroup$
Correction - should have read $ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{dx}{dt} + frac{partial z}{partial y} cdot frac{dy}{dt}$
$endgroup$
– Buddy
Sep 28 '16 at 2:00






$begingroup$
Correction - should have read $ frac{dz}{dt} =frac{partial z}{partial x} cdot frac{dx}{dt} + frac{partial z}{partial y} cdot frac{dy}{dt}$
$endgroup$
– Buddy
Sep 28 '16 at 2:00












1 Answer
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$begingroup$

We need to find:



$$text{z}'(t)=frac{text{d}text{z}(t)}{text{d}t}=frac{text{d}}{text{d}t}left(e^{-3t}cdotleft(-sin(2t)right)^2right)=$$
$$frac{text{d}}{text{d}t}left(e^{-3t}sin^2(2t)right)=e^{-3t}sin(2t)left(4cos(2t)-3sin(2t)right)$$




Using:




  1. The product rule:
    $$frac{text{d}}{text{d}t}left(f(t)cdot y(t)right)=f(t)cdotfrac{text{d}}{text{d}t}left(y(t)right)+y(t)cdotfrac{text{d}}{text{d}t}left(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$

  2. $$frac{text{d}}{text{d}t}left(e^{x(t)}right)=e^{x(t)}cdotfrac{text{d}}{text{d}t}left(x(t)right)=x'(t)cdot e^{x(t)}$$

  3. When $text{C}$ is a constant:
    $$frac{text{d}}{text{d}t}left(text{C}cdot q(t)right)=text{C}cdotfrac{text{d}}{text{d}t}left(q(t)right)=text{C}cdot q'(t)$$

  4. When $text{n}$ is a constant, using the chain rule:
    $$frac{text{d}}{text{d}t}left(w(t)^text{n}right)=text{n}cdot w(t)^{text{n}-1}cdotfrac{text{d}}{text{d}t}left(w(t)right)=text{n}cdot w(t)^{text{n}-1}cdot w'(t)$$

  5. When $text{m}$ is a constant, using the chain rule:
    $$frac{text{d}}{text{d}t}left(v(text{m}t)right)=frac{text{d}}{text{d}t}left(text{m}tright)cdot v'(text{m}t)=text{m}cdot v'(text{m}t)$$






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    1 Answer
    1






    active

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    active

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    0












    $begingroup$

    We need to find:



    $$text{z}'(t)=frac{text{d}text{z}(t)}{text{d}t}=frac{text{d}}{text{d}t}left(e^{-3t}cdotleft(-sin(2t)right)^2right)=$$
    $$frac{text{d}}{text{d}t}left(e^{-3t}sin^2(2t)right)=e^{-3t}sin(2t)left(4cos(2t)-3sin(2t)right)$$




    Using:




    1. The product rule:
      $$frac{text{d}}{text{d}t}left(f(t)cdot y(t)right)=f(t)cdotfrac{text{d}}{text{d}t}left(y(t)right)+y(t)cdotfrac{text{d}}{text{d}t}left(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$

    2. $$frac{text{d}}{text{d}t}left(e^{x(t)}right)=e^{x(t)}cdotfrac{text{d}}{text{d}t}left(x(t)right)=x'(t)cdot e^{x(t)}$$

    3. When $text{C}$ is a constant:
      $$frac{text{d}}{text{d}t}left(text{C}cdot q(t)right)=text{C}cdotfrac{text{d}}{text{d}t}left(q(t)right)=text{C}cdot q'(t)$$

    4. When $text{n}$ is a constant, using the chain rule:
      $$frac{text{d}}{text{d}t}left(w(t)^text{n}right)=text{n}cdot w(t)^{text{n}-1}cdotfrac{text{d}}{text{d}t}left(w(t)right)=text{n}cdot w(t)^{text{n}-1}cdot w'(t)$$

    5. When $text{m}$ is a constant, using the chain rule:
      $$frac{text{d}}{text{d}t}left(v(text{m}t)right)=frac{text{d}}{text{d}t}left(text{m}tright)cdot v'(text{m}t)=text{m}cdot v'(text{m}t)$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We need to find:



      $$text{z}'(t)=frac{text{d}text{z}(t)}{text{d}t}=frac{text{d}}{text{d}t}left(e^{-3t}cdotleft(-sin(2t)right)^2right)=$$
      $$frac{text{d}}{text{d}t}left(e^{-3t}sin^2(2t)right)=e^{-3t}sin(2t)left(4cos(2t)-3sin(2t)right)$$




      Using:




      1. The product rule:
        $$frac{text{d}}{text{d}t}left(f(t)cdot y(t)right)=f(t)cdotfrac{text{d}}{text{d}t}left(y(t)right)+y(t)cdotfrac{text{d}}{text{d}t}left(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$

      2. $$frac{text{d}}{text{d}t}left(e^{x(t)}right)=e^{x(t)}cdotfrac{text{d}}{text{d}t}left(x(t)right)=x'(t)cdot e^{x(t)}$$

      3. When $text{C}$ is a constant:
        $$frac{text{d}}{text{d}t}left(text{C}cdot q(t)right)=text{C}cdotfrac{text{d}}{text{d}t}left(q(t)right)=text{C}cdot q'(t)$$

      4. When $text{n}$ is a constant, using the chain rule:
        $$frac{text{d}}{text{d}t}left(w(t)^text{n}right)=text{n}cdot w(t)^{text{n}-1}cdotfrac{text{d}}{text{d}t}left(w(t)right)=text{n}cdot w(t)^{text{n}-1}cdot w'(t)$$

      5. When $text{m}$ is a constant, using the chain rule:
        $$frac{text{d}}{text{d}t}left(v(text{m}t)right)=frac{text{d}}{text{d}t}left(text{m}tright)cdot v'(text{m}t)=text{m}cdot v'(text{m}t)$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We need to find:



        $$text{z}'(t)=frac{text{d}text{z}(t)}{text{d}t}=frac{text{d}}{text{d}t}left(e^{-3t}cdotleft(-sin(2t)right)^2right)=$$
        $$frac{text{d}}{text{d}t}left(e^{-3t}sin^2(2t)right)=e^{-3t}sin(2t)left(4cos(2t)-3sin(2t)right)$$




        Using:




        1. The product rule:
          $$frac{text{d}}{text{d}t}left(f(t)cdot y(t)right)=f(t)cdotfrac{text{d}}{text{d}t}left(y(t)right)+y(t)cdotfrac{text{d}}{text{d}t}left(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$

        2. $$frac{text{d}}{text{d}t}left(e^{x(t)}right)=e^{x(t)}cdotfrac{text{d}}{text{d}t}left(x(t)right)=x'(t)cdot e^{x(t)}$$

        3. When $text{C}$ is a constant:
          $$frac{text{d}}{text{d}t}left(text{C}cdot q(t)right)=text{C}cdotfrac{text{d}}{text{d}t}left(q(t)right)=text{C}cdot q'(t)$$

        4. When $text{n}$ is a constant, using the chain rule:
          $$frac{text{d}}{text{d}t}left(w(t)^text{n}right)=text{n}cdot w(t)^{text{n}-1}cdotfrac{text{d}}{text{d}t}left(w(t)right)=text{n}cdot w(t)^{text{n}-1}cdot w'(t)$$

        5. When $text{m}$ is a constant, using the chain rule:
          $$frac{text{d}}{text{d}t}left(v(text{m}t)right)=frac{text{d}}{text{d}t}left(text{m}tright)cdot v'(text{m}t)=text{m}cdot v'(text{m}t)$$






        share|cite|improve this answer









        $endgroup$



        We need to find:



        $$text{z}'(t)=frac{text{d}text{z}(t)}{text{d}t}=frac{text{d}}{text{d}t}left(e^{-3t}cdotleft(-sin(2t)right)^2right)=$$
        $$frac{text{d}}{text{d}t}left(e^{-3t}sin^2(2t)right)=e^{-3t}sin(2t)left(4cos(2t)-3sin(2t)right)$$




        Using:




        1. The product rule:
          $$frac{text{d}}{text{d}t}left(f(t)cdot y(t)right)=f(t)cdotfrac{text{d}}{text{d}t}left(y(t)right)+y(t)cdotfrac{text{d}}{text{d}t}left(f(t)right)=y(t)cdot f'(t)+f(t)cdot y'(t)$$

        2. $$frac{text{d}}{text{d}t}left(e^{x(t)}right)=e^{x(t)}cdotfrac{text{d}}{text{d}t}left(x(t)right)=x'(t)cdot e^{x(t)}$$

        3. When $text{C}$ is a constant:
          $$frac{text{d}}{text{d}t}left(text{C}cdot q(t)right)=text{C}cdotfrac{text{d}}{text{d}t}left(q(t)right)=text{C}cdot q'(t)$$

        4. When $text{n}$ is a constant, using the chain rule:
          $$frac{text{d}}{text{d}t}left(w(t)^text{n}right)=text{n}cdot w(t)^{text{n}-1}cdotfrac{text{d}}{text{d}t}left(w(t)right)=text{n}cdot w(t)^{text{n}-1}cdot w'(t)$$

        5. When $text{m}$ is a constant, using the chain rule:
          $$frac{text{d}}{text{d}t}left(v(text{m}t)right)=frac{text{d}}{text{d}t}left(text{m}tright)cdot v'(text{m}t)=text{m}cdot v'(text{m}t)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 28 '16 at 19:54









        JanJan

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        21.7k31240






























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