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,

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

We have

$w=7ye^x-\ln z\implies\begin{cases}\dfrac{\partial w}{\partial x}=7ye^x\\\\\dfrac{\partial w}{\partial y}=7e^x\\\\\dfrac{\partial w}{\partial z}=-\dfrac1z\end{cases}$

and

$\begin{cases}x=\ln(t^2+1)\\y=\tan^{-1}t\\z=e^t\end{cases}\implies\begin{cases}\dfrac{\mathrm dx}{\mathrm dt}=\dfrac{2t}{t^2+1}\\\\\dfrac{\mathrm dy}{\mathrm dt}=\dfrac1{t^2+1}\\\\\dfrac{\mathrm dz}{\mathrm dt}=e^t\end{cases}$

Putting everything together, we get

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{14ye^xt}{t^2+1}+\dfrac{7e^x}{t^2+1}-\dfrac{e^t}z$

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

$\dfrac{\mathrm dw}{\mathrm dt}=14yt+7-1$

$\dfrac{\mathrm dw}{\mathrm dt}=14t\tan^{-1}t+6$

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