Study23 3 years ago Find an equation of the tangent curve at the given point? $$\ \large y=sin(sinx), (\pi,0)$$.

1. ash2326

@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

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

3. ash2326

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

4. Study23

Okay, one moment while I find the derivative

5. ash2326

ok

6. Study23

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

7. Study23

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

8. ash2326

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

9. Study23

? 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

you'd get $\cos (\sin x)\times \cos x$

11. Study23

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

12. ash2326

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

13. Study23

Okay

14. Study23

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