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Calculus1 14 Online
OpenStudy (anonymous):

Find dy/dt given that [y=u^2+4, u=sin(x), (pi)t]

OpenStudy (dumbcow):

\[\frac{dy}{dt} = \frac{dy}{du}*\frac{du}{dt}\]

hartnn (hartnn):

i think the Q has x=pi*t so, \(\large \dfrac{dy}{dt} = \dfrac{dy}{du}*\dfrac{du}{dx}*\dfrac{dx}{dt}\) and are you able to post something ? i see you were trying to write but nothing showed up.....if you can't you can message us...

OpenStudy (anonymous):

Yes I am able to post, just can't think of how to word my question

OpenStudy (anonymous):

For the derivative of y, would I plug in the value of u and then x? so i would be finding \[\sin(\pi(t))^2\]

OpenStudy (anonymous):

sorry, i forgot the +4 after the equation

hartnn (hartnn):

you can do that and then apply thye chain rule, but the simpler way is to use the chain rule formula we gave you, so you will just need to calculate dy/du from y=u^2+4 du/dxfrom u=sin x and dx/dt from x=pi*t which is much simpler :)

OpenStudy (anonymous):

After calculating the derovatives, I came out with 2u(cos(x))pi. After this I just multiply across and that will be my answer?

hartnn (hartnn):

you will need to put the final answer in terms of t so, first plug in x=pi, then u= sin x = sin pi*t

hartnn (hartnn):

***x=pi*t

OpenStudy (anonymous):

OOOOH ok thank you so much

hartnn (hartnn):

welcome ^_^

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