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Study23 Group Title

Find an equation of the tangent curve at the given point? \(\ \large y=sin(sinx), (\pi,0) \).

  • one year ago
  • one year ago

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  1. ash2326 Group Title
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    @Study23 Slope of the tangent at the point \((\pi, 0)\) to the curve \(y=\sin (\sin x)\) is given by \[\frac{dy}{dx}_{(\pi, 0)}\] Can you find ? \[\frac{dy}{dx}\]

    • one year ago
  2. Study23 Group Title
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    Could you help me with that? I'm not sure how to use the values to find the derivative, @ash2326

    • one year ago
  3. ash2326 Group Title
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    Just find the derivative, we'll use the values later.

    • one year ago
  4. Study23 Group Title
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    Okay, one moment while I find the derivative

    • one year ago
  5. ash2326 Group Title
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    ok

    • one year ago
  6. Study23 Group Title
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    \(\ \Huge \text{So, I got: } y'=cos^2x \)

    • one year ago
  7. Study23 Group Title
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    \(\ \large \text{How do I proceed from here?} \)

    • one year ago
  8. ash2326 Group Title
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    \[\frac{dy}{dx}=\frac{d}{dx} (\sin (\sin x))\] \[\frac{dy}{dx}=\cos (\sin x)\times\frac{d}{dx} \sin (x)\]

    • one year ago
  9. Study23 Group Title
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    ? What I did: \(\ \large y=sin(sinx) \) \(\ \large y'=cos\frac{d}{dx}(sinx)\) \(\ \large y' = cos \times cosx,\) \(\ \large \text{So, } cos^2x \)

    • one year ago
  10. ash2326 Group Title
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    you'd get \[\cos (\sin x)\times \cos x\]

    • one year ago
  11. Study23 Group Title
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    Okay, I got that this time... So now what?

    • one year ago
  12. ash2326 Group Title
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    now put x=\(\pi\) here to find the slope of the tangent

    • one year ago
  13. Study23 Group Title
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    Okay

    • one year ago
  14. Study23 Group Title
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    So, -1*-1=1? That's the slope?

    • one year ago
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