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anonymous
 one year ago
If f(x)=integral ^sinx, o on the bottom, root1+t^2 dt and g(y)=integral^y, 3 on the bottom f(x) dx find g''(pi/6) my answer is root15/4 is this correct?
anonymous
 one year ago
If f(x)=integral ^sinx, o on the bottom, root1+t^2 dt and g(y)=integral^y, 3 on the bottom f(x) dx find g''(pi/6) my answer is root15/4 is this correct?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(x) = \int_0^{\sin x}\sqrt{1+t^2}~dt \]And\[ g(x) = \int_3^y f(x)~dx \]Okay, how did you get your answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ f(x) = \int_0^{\sin x}\sqrt{1+t^2}~dt \]And\[ g(x) = \int_3^y f(x)~dx \]Okay, how did you get your answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i did root1+(sin pi/6)^2 * cos(pi/6)= root1+1/4 + root3/2= root5/4 * root3/2= root15/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The issue is, don't we need \(y=h(x)\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's why I'm a bit confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait, let me think.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ g(y) = \int_3^y f(x)~dx \]So then:\[ g'(y) = f(y) \]And:\[ g''(y) = f'(y) \]Then \[ f(y) =\int_0^{\sin y} \sqrt{1+t^2}~dt \]So\[ f'(y) = (\cos y)\sqrt{1+\sin^2 y} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you are right.
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