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Brooke_army Group TitleBest ResponseYou've already chosen the best response.0
ok for this question i think im doing something wrong i got the answer of 17 but this is not right please help
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
can you find dy/dx of the given equation?
 one year ago

Brooke_army Group TitleBest ResponseYou've already chosen the best response.0
dy/dt=3x^2+5
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
when you take the derivative wrt to t you should get a dx/dt term
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
u need to use chain rule d/dt(5x) = 5 d/dx (dx/dt)
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
instead of d/dx x^3 = 3 x^2 dx/dx you do d/dt x^3 = 3 x^2 dx/dt
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
if you think \[ \frac{d}{dx} x= 1\] think this \[ \frac{ d}{dx} x= \frac{dx}{dx}= 1\]
 one year ago

sunnymony Group TitleBest ResponseYou've already chosen the best response.0
dw:1352559750685:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
using the same logic \[ \frac{d }{dt} x= \frac{dx}{dt} \]
 one year ago

sunnymony Group TitleBest ResponseYou've already chosen the best response.0
3x2+5 is ans
 one year ago

Brooke_army Group TitleBest ResponseYou've already chosen the best response.0
thanks so much i got the wright answer!!
 one year ago

sunnymony Group TitleBest ResponseYou've already chosen the best response.0
dy/dx= 3x2+5 dx/dt = 5 so both multiply we get =15x2+25
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
then you got dy/dt=3x^2 dx/dt + 5 dx/dt replace x=2, dx/dt=5 to get dy/dt = 3*4*5+ 5*5= 85
 one year ago

sunnymony Group TitleBest ResponseYou've already chosen the best response.0
so 15( 2x2)+25 =60+25 85
 one year ago

sunnymony Group TitleBest ResponseYou've already chosen the best response.0
thats right @phi
 one year ago
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