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Study23

  • 2 years ago

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

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  1. ash2326
    • 2 years ago
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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 ? \[\frac{dy}{dx}\]

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

  3. ash2326
    • 2 years ago
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    Just find the derivative, we'll use the values later.

  4. Study23
    • 2 years ago
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    Okay, one moment while I find the derivative

  5. ash2326
    • 2 years ago
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    ok

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

  7. Study23
    • 2 years ago
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    \(\ \large \text{How do I proceed from here?} \)

  8. ash2326
    • 2 years ago
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    \[\frac{dy}{dx}=\frac{d}{dx} (\sin (\sin x))\] \[\frac{dy}{dx}=\cos (\sin x)\times\frac{d}{dx} \sin (x)\]

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

  10. ash2326
    • 2 years ago
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    you'd get \[\cos (\sin x)\times \cos x\]

  11. Study23
    • 2 years ago
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    Okay, I got that this time... So now what?

  12. ash2326
    • 2 years ago
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    now put x=\(\pi\) here to find the slope of the tangent

  13. Study23
    • 2 years ago
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    Okay

  14. Study23
    • 2 years ago
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    So, -1*-1=1? That's the slope?

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