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

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ash2326
 2 years ago
Best ResponseYou've already chosen the best response.0@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 ? \[\frac{dy}{dx}\]

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0Could you help me with that? I'm not sure how to use the values to find the derivative, @ash2326

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.0Just find the derivative, we'll use the values later.

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, one moment while I find the derivative

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0\(\ \Huge \text{So, I got: } y'=cos^2x \)

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0\(\ \large \text{How do I proceed from here?} \)

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx}=\frac{d}{dx} (\sin (\sin x))\] \[\frac{dy}{dx}=\cos (\sin x)\times\frac{d}{dx} \sin (x)\]

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0? What I did: \(\ \large y=sin(sinx) \) \(\ \large y'=cos\frac{d}{dx}(sinx)\) \(\ \large y' = cos \times cosx,\) \(\ \large \text{So, } cos^2x \)

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.0you'd get \[\cos (\sin x)\times \cos x\]

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, I got that this time... So now what?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.0now put x=\(\pi\) here to find the slope of the tangent

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0So, 1*1=1? That's the slope?
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