please help, explanation would be greatly appreciated (picture included)

- anonymous

please help, explanation would be greatly appreciated (picture included)

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- anonymous

##### 1 Attachment

- DanJS

y=2*e^cos(x)
\[\frac{ dy }{ dt } = 5 unit/s\]

- DanJS

do a d/dt to both sides of that equation

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## More answers

- anonymous

to which equation?

- anonymous

to dy/d=2(lne)(e^cosx)(-sinx)

- anonymous

i mean that that is the derivative of y=2e^cosx

- anonymous

im sorry but i dont know what that means

- DanJS

y is a function of x is a function of t, y--->x--->t

- DanJS

y ( x(t))

- anonymous

would t be the 5 units per second?

- anonymous

no nvm i know that wrong

- DanJS

|dw:1449434843192:dw|

- DanJS

to get from y to the t, you do dy/dx * dx/dt

- DanJS

notice that is dy/dt = (dy/dx)*(dx/dt) -- both sides same

- anonymous

alright what do we do next?

- DanJS

the thing is compounded like that
\[y(x[t]) = 2e^{\cos[(x(t)]}\]

- anonymous

and we need to find the derivative of that equation

- DanJS

they give you some info
dy/dx = 5 unit/s always
x = pi/2
they want you to find dx/dt at the instant when the angle x(t) is pi/2

- anonymous

i dont understadn how i would do that

- DanJS

*dy/dt = 5 mistype

- anonymous

i still dont get it im very sorry

- DanJS

dy/dt = dy/dx * dx/dt
take derivative w.r.t. x first, then w.r.t t

- anonymous

whst is w.r.t

- DanJS

with respect to,

- DanJS

dy/dt = dy/dx * dx/dt

- DanJS

derivative of y w.r.t.x times derivative of x w.r.t t

- anonymous

how am i supposed to do that if i dont know what dy/dx is or dx/dt

- anonymous

all i know is that dy/dt is 5 units/sec

- DanJS

like remember doing chain rule derivatives... you have to multiply by the derivative of the inner function
ex
[ d/dx] * cos(x^2+2) = -sin(x^2+2) * ( 2x)

- DanJS

anyway here...

- anonymous

thanks for trying but i have somewhere i be in an hour and i have to leave now, if i wasn't in a hurry i could probably understand it better thank you

- DanJS

\[\large y = 2e^{\cos(x)}\]
derivative with respect to x, d/dx both sides

- DanJS

\[\large \frac{ dy }{ dx } = -2*\sin(x) *e^{\cos(x)}\]

- DanJS

a step further realizing xis a function of t, the chain rule gives you dy/dt
\[\huge \frac{ d }{d t }=\frac{ dy }{ dx }*\frac{ dx }{ dt }\]

- anonymous

but thats what i said was the derivative at the beginning!!

- anonymous

so we just solve for dx/dt next?

- DanJS

no actual x(t) function is said, so just add that dx/dt on the end
\[\frac{ dy }{ dt } = \large \frac{ dy }{ dx } * \frac{ dx }{ dt } = -2*\sin(x) *e^{\cos(x)} * \frac{ dx }{ dt }\]

- DanJS

use that to figure the rate x is changing with the time, dx/dt, when x is that given angle
5 = -2*sin(x)*e^(cos(x)) * [dx/dt]

- DanJS

\[5 = -2*\sin(\pi/2)*e^{\cos(\pi/2)}* \frac{ dx }{ dt }\]

- DanJS

5 = -2 * 1 * 1 * dx/dt
dx/dt = 5/-2 unit/sec

- DanJS

just need to practice the chain rule thing for nested functions
to get dy/dt
you have to do d/dx first and multiply by dx/dt

- DanJS

like A [ B [ C [ D(x) ] ] ]
same, to get to dA/dx there, you need
dA/dB * dB/dC * dC/dD * dD/dx = dA*/Dx

- DanJS

gl

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