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Find an equation of the tangent curve at the given point?
\(\ \large y=sin(sinx), (\pi,0) \).
 one year ago
 one year ago
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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ash2326Best 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 ? \[\frac{dy}{dx}\]
 one year ago

Study23Best ResponseYou've already chosen the best response.0
Could you help me with that? I'm not sure how to use the values to find the derivative, @ash2326
 one year ago

ash2326Best ResponseYou've already chosen the best response.0
Just find the derivative, we'll use the values later.
 one year ago

Study23Best ResponseYou've already chosen the best response.0
Okay, one moment while I find the derivative
 one year ago

Study23Best ResponseYou've already chosen the best response.0
\(\ \Huge \text{So, I got: } y'=cos^2x \)
 one year ago

Study23Best ResponseYou've already chosen the best response.0
\(\ \large \text{How do I proceed from here?} \)
 one year ago

ash2326Best ResponseYou've already chosen the best response.0
\[\frac{dy}{dx}=\frac{d}{dx} (\sin (\sin x))\] \[\frac{dy}{dx}=\cos (\sin x)\times\frac{d}{dx} \sin (x)\]
 one year ago

Study23Best ResponseYou've already chosen the best response.0
? What I did: \(\ \large y=sin(sinx) \) \(\ \large y'=cos\frac{d}{dx}(sinx)\) \(\ \large y' = cos \times cosx,\) \(\ \large \text{So, } cos^2x \)
 one year ago

ash2326Best ResponseYou've already chosen the best response.0
you'd get \[\cos (\sin x)\times \cos x\]
 one year ago

Study23Best ResponseYou've already chosen the best response.0
Okay, I got that this time... So now what?
 one year ago

ash2326Best ResponseYou've already chosen the best response.0
now put x=\(\pi\) here to find the slope of the tangent
 one year ago

Study23Best ResponseYou've already chosen the best response.0
So, 1*1=1? That's the slope?
 one year ago
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