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

Mathematics
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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 ? \[\frac{dy}{dx}\]
Could you help me with that? I'm not sure how to use the values to find the derivative, @ash2326
Just find the derivative, we'll use the values later.

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Okay, one moment while I find the derivative
ok
\(\ \Huge \text{So, I got: } y'=cos^2x \)
\(\ \large \text{How do I proceed from here?} \)
\[\frac{dy}{dx}=\frac{d}{dx} (\sin (\sin x))\] \[\frac{dy}{dx}=\cos (\sin x)\times\frac{d}{dx} \sin (x)\]
? What I did: \(\ \large y=sin(sinx) \) \(\ \large y'=cos\frac{d}{dx}(sinx)\) \(\ \large y' = cos \times cosx,\) \(\ \large \text{So, } cos^2x \)
you'd get \[\cos (\sin x)\times \cos x\]
Okay, I got that this time... So now what?
now put x=\(\pi\) here to find the slope of the tangent
Okay
So, -1*-1=1? That's the slope?

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