Calc2 Finding $f'(2)$ from tangent line












2












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So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$



Any ideas on how to go about this? Thank you!






derivatives tangent-line







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    Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
    $endgroup$
    – The Chaz 2.0
    Jan 25 at 3:09
















2












$begingroup$


So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$



Any ideas on how to go about this? Thank you!






derivatives tangent-line







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
    $endgroup$
    – The Chaz 2.0
    Jan 25 at 3:09














2












2








2





$begingroup$


So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$



Any ideas on how to go about this? Thank you!






derivatives tangent-line







share|cite|improve this question











$endgroup$




So, I have a Calc 2 problem I am stuck on. The tangent line to $h(x)$ at $x = 2$ is $3x - 2$. It says to find $f'(2)$ given $f(x) = -3[h(x)]^2 + 2x + 2$



Any ideas on how to go about this? Thank you!






derivatives tangent-line




derivatives tangent-line






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 3:16









Christopher Marley

1,129115




1,129115










asked Jan 25 at 3:05









Jelly BiscuitJelly Biscuit

204




204








  • 1




    $begingroup$
    Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
    $endgroup$
    – The Chaz 2.0
    Jan 25 at 3:09














  • 1




    $begingroup$
    Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
    $endgroup$
    – The Chaz 2.0
    Jan 25 at 3:09








1




1




$begingroup$
Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
$endgroup$
– The Chaz 2.0
Jan 25 at 3:09




$begingroup$
Take the derivative, use the chain rule, observe that $h(2) = 4$, etc.
$endgroup$
– The Chaz 2.0
Jan 25 at 3:09










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.



Now, $$f(x) = -3[h(x)]^2+2x+2$$
By chain rule for h(x)
$$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
x=-2
$$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
Simplify
$$f'(-2) = -6cdot 4cdot 3+2 = -70$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you so much!
    $endgroup$
    – Jelly Biscuit
    Jan 25 at 5:46



















2












$begingroup$

We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $






share|cite|improve this answer











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    2 Answers
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    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.



    Now, $$f(x) = -3[h(x)]^2+2x+2$$
    By chain rule for h(x)
    $$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
    x=-2
    $$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
    Simplify
    $$f'(-2) = -6cdot 4cdot 3+2 = -70$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      thank you so much!
      $endgroup$
      – Jelly Biscuit
      Jan 25 at 5:46
















    1












    $begingroup$

    Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.



    Now, $$f(x) = -3[h(x)]^2+2x+2$$
    By chain rule for h(x)
    $$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
    x=-2
    $$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
    Simplify
    $$f'(-2) = -6cdot 4cdot 3+2 = -70$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      thank you so much!
      $endgroup$
      – Jelly Biscuit
      Jan 25 at 5:46














    1












    1








    1





    $begingroup$

    Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.



    Now, $$f(x) = -3[h(x)]^2+2x+2$$
    By chain rule for h(x)
    $$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
    x=-2
    $$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
    Simplify
    $$f'(-2) = -6cdot 4cdot 3+2 = -70$$






    share|cite|improve this answer











    $endgroup$



    Let's compile information: The slope of the tangent line at a point equals the derivative. So $h'(2) = 3$. Secondly, the tangent line touches the function at its value, so $h(2) = 3cdot2-2 = 4$.



    Now, $$f(x) = -3[h(x)]^2+2x+2$$
    By chain rule for h(x)
    $$f'(x) = -3cdot2cdot[h(x)]cdot [h'(x)]+2$$
    x=-2
    $$f'(-2) = -6cdot [h(-2)]*[h'(-2)]+2$$
    Simplify
    $$f'(-2) = -6cdot 4cdot 3+2 = -70$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 25 at 4:13









    El borito

    666216




    666216










    answered Jan 25 at 3:13









    Christopher MarleyChristopher Marley

    1,129115




    1,129115












    • $begingroup$
      thank you so much!
      $endgroup$
      – Jelly Biscuit
      Jan 25 at 5:46


















    • $begingroup$
      thank you so much!
      $endgroup$
      – Jelly Biscuit
      Jan 25 at 5:46
















    $begingroup$
    thank you so much!
    $endgroup$
    – Jelly Biscuit
    Jan 25 at 5:46




    $begingroup$
    thank you so much!
    $endgroup$
    – Jelly Biscuit
    Jan 25 at 5:46











    2












    $begingroup$

    We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $






        share|cite|improve this answer











        $endgroup$



        We are given $h'(2)=3$ and since $f'(x) = - 6hh' + 2 $ and since $h(2)=4$ one obtains that $f'(2) = -6 cdot 4 cdot 3 + 2 = boxed{-70} $







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 25 at 3:16

























        answered Jan 25 at 3:10









        Jimmy SabaterJimmy Sabater

        3,023325




        3,023325






























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