Question

For the given functions, (a) express dw/dt as a function of t, both by using the chain rule and by expressing w in terms of t and differentiating directly with respect to to Then (b) evaluate dw/dt at the given value of t. w = 7y e^x - Inz,x = ln(t^2 + 1), y = tan^-1t,z=e^t;t = 1

(a) dw/dt =?

Answer 1

By the chain rule,

`(\mathrm dw)/(\mathrm dt)=(\partial w)/(\partial x)(\mathrm dx)/(\mathrm dt)+(\partial w)/(\partial y)(\mathrm dy)/(\mathrm dt)+(\partial w)/(\partial z)(\mathrm dz)/(\mathrm dt)`

We have

`w=7ye^x-\ln z\implies\begin{cases}(\partial w)/(\partial x)=7ye^x\\\\(\partial w)/(\partial y)=7e^x\\\\(\partial w)/(\partial z)=-\frac1z\end{cases}`

and

`\begin{cases}x=\ln(t^2+1)\\y=\tan^(-1)t\\z=e^t\end{cases}\implies\begin{cases}(\mathrm dx)/(\mathrm dt)=(2t)/(t^2+1)\\\\(\mathrm dy)/(\mathrm dt)=\frac1{t^2+1}\\\\(\mathrm dz)/(\mathrm dt)=e^t\end{cases}`

Putting everything together, we get

`(\mathrm dw)/(\mathrm dt)=(14ye^xt)/(t^2+1)+(7e^x)/(t^2+1)-\frac{e^t}z`

`x=\ln(t^2+1)`, so
`e^x=e^(\ln(t^2+1))=t^2+1`, and
`z=e^t`, so
`\frac{e^t}z=1`.

`(\mathrm dw)/(\mathrm dt)=14yt+7-1`

`(\mathrm dw)/(\mathrm dt)=14t\tan^(-1)t+6`

Then when
`t=1`, the derivative has a value of
`\frac{7\pi}2+6`.