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












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












$begingroup$


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






share|cite|improve this question















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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086655%2fcalc2-finding-f2-from-tangent-line%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086655%2fcalc2-finding-f2-from-tangent-line%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            'app-layout' is not a known element: how to share Component with different Modules

            Image
            0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
            Read more

            android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
            Read more

            WPF add header to Image with URL pettitions [duplicate]

            Image
            up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
            Read more