Question

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

Answer 1

@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 ?
\[\frac{dy}{dx}\]

Answer 2

Could you help me with that? I'm not sure how to use the values to find the derivative, @ash2326

Answer 3

Just find the derivative, we'll use the values later.

Answer 4

Okay, one moment while I find the derivative

Answer 5

ok

Answer 6

\(\ \Huge \text{So, I got: } y'=cos^2x \)

Answer 7

\(\ \large \text{How do I proceed from here?} \)

Answer 8

\[\frac{dy}{dx}=\frac{d}{dx} (\sin (\sin x))\]
\[\frac{dy}{dx}=\cos (\sin x)\times\frac{d}{dx} \sin (x)\]

Answer 9

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

Answer 10

you'd get
\[\cos (\sin x)\times \cos x\]

Answer 11

Okay, I got that this time... So now what?

Answer 12

now put x=\(\pi\) here to find the slope of the tangent

Answer 13

Okay

Answer 14

So, -1*-1=1? That's the slope?