Question

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

Answer 1

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.