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anonymous
 5 years ago
Use the Chain Rule to find dw/dt. (Enter your answer only in terms of t.) w = xe^(y /z), x = t^6, y = 9  t, z = 4 + 4t
anonymous
 5 years ago
Use the Chain Rule to find dw/dt. (Enter your answer only in terms of t.) w = xe^(y /z), x = t^6, y = 9  t, z = 4 + 4t

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, it would be best to substitute the variables in to get w as a function of t. This may be messy, but it makes it easier to see and you won't have to do a lot of implicit differentiation. So, the equation with everything subbed in is: \[w=t^{6}e^{(9t)/(4+4t)}\] Now in order to take the derivative with respect to t, you must do the product rule. The product rule of 2 functions is: \[d/dt(u(t)v(t))=u'(t)v(t) + u(t)v'(t)\] where u(t) = t^6 and v(t) = e^(9t)/(4+4t) Now the derivative dw/dt = \[6t^{5}e^{(9t)/(4+4t)} + t^{6}(d/dt)(e^{(9t)/(4+4t)})\] Now, the derivative of \[e^{(9t)/(4+4t)}\] with respect to t must be done using the chain rule, which can be said as follows: The derivative of the outer function with the inner function plugged in times the derivative of the inner function. The outer function is \[e^{t}\] and the inner function is \[(9−t)/(4+4t)\] The derivative of the outer of \[e^{t}\] is \[e^{t}\] plug the inner function in for t to get: \[e^{(9t)/(4+4t)}\] Multiply the above by the derivative of \[(9−t)/(4+4t)\] which must be done using the quotient rule or: \[d/dt(u(t)/v(t))=(u'(t)v(t)u(t)v'(t))/v^{2}(t)\] where u(t) = 9  t and v(t) = 4 + 4t Thus the derivative is: \[((1)(4+4t)  (9t)(4))/(4+4t)^{2}\] So, the derivative of \[e^{(9−t)/(4+4t)}\] is \[e^{(9−t)/(4+4t)} * ((1)(4+4t)  (9t)(4))/(4+4t)^{2}\] making \[dw/dt=6t^{5}e^{(9t)/(4+4t)} +\] \[ t^{6}*e^{(9t)/(4+4t)}*((4+4t)4(9t))/(4+4t)^{2}\] Now just simplify.
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