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BlackZzzverrR [31]
3 years ago
14

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 an

d 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 =?
Mathematics
1 answer:
ch4aika [34]3 years ago
6 0

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.

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