anonymous
  • anonymous
Find an equation of the tangent curve at the given point? \(\ \large y=sin(sinx), (\pi,0) \).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ash2326
  • 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 ? \[\frac{dy}{dx}\]
anonymous
  • anonymous
Could you help me with that? I'm not sure how to use the values to find the derivative, @ash2326
ash2326
  • ash2326
Just find the derivative, we'll use the values later.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Okay, one moment while I find the derivative
ash2326
  • ash2326
ok
anonymous
  • anonymous
\(\ \Huge \text{So, I got: } y'=cos^2x \)
anonymous
  • anonymous
\(\ \large \text{How do I proceed from here?} \)
ash2326
  • ash2326
\[\frac{dy}{dx}=\frac{d}{dx} (\sin (\sin x))\] \[\frac{dy}{dx}=\cos (\sin x)\times\frac{d}{dx} \sin (x)\]
anonymous
  • anonymous
? 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
  • ash2326
you'd get \[\cos (\sin x)\times \cos x\]
anonymous
  • anonymous
Okay, I got that this time... So now what?
ash2326
  • ash2326
now put x=\(\pi\) here to find the slope of the tangent
anonymous
  • anonymous
Okay
anonymous
  • anonymous
So, -1*-1=1? That's the slope?

Looking for something else?

Not the answer you are looking for? Search for more explanations.