anonymous
  • anonymous
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
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Okay, 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: \[w=t^{6}e^{(9-t)/(4+4t)}\] 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^(9-t)/(4+4t) Now the derivative dw/dt = \[6t^{5}e^{(9-t)/(4+4t)} + t^{6}(d/dt)(e^{(9-t)/(4+4t)})\] Now, the derivative of \[e^{(9-t)/(4+4t)}\] 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: \[e^{(9-t)/(4+4t)}\] 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: \[((-1)(4+4t) - (9-t)(4))/(4+4t)^{2}\] So, the derivative of \[e^{(9−t)/(4+4t)}\] is \[e^{(9−t)/(4+4t)} * ((-1)(4+4t) - (9-t)(4))/(4+4t)^{2}\] making \[dw/dt=6t^{5}e^{(9-t)/(4+4t)} +\] \[ t^{6}*e^{(9-t)/(4+4t)}*(-(4+4t)-4(9-t))/(4+4t)^{2}\] Now just simplify.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.