Short Answer

Determine the derivative \(\frac{{dz}}{{dt}}\)at the given value of \(t.\)The functions are \(z = f(x,y),x = g(t)\) and \({\rm{ }}y = h(t){\rm{.}}\)

The derivative of \(\frac{{dz}}{{dt}}\) at \(t = 3\)is \(62.\)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The chain rule.

With the chain rule, the partial derivative of function \({\rm{f(x, y)}}\) with respect to \({\rm{t}}\) becomes \(\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}.\)

Step 2: Use the chain rule for calculation.

The given functions are \(z = f(x,y),x = g(t)\) and \({\rm{ }}y = h(t).\)

Differentiate \(z\) with respect to \(t\)as follows.

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial t}} = \frac{\partial }{{\partial t}}f(x,y)\\ = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}\\ = {f_x}(g(t),h(t)) \cdot \frac{{dg}}{{dt}} + {f_y}(g(t),h(t)) \cdot \frac{{dh}}{{dt}}\end{aligned}\)

At\(t = 3\), the partial derivative becomes.

\(\begin{aligned}{l}{\left. {\frac{{\partial z}}{{\partial t}}} \right|_{t = 3}} = {f_x}(g(3),h(3)) \cdot \frac{{dg}}{{dt}}(3) + {f_y}(g(3),h(3)) \cdot \frac{{dh}}{{dt}}(3)\\ = {f_x}(2,7) \cdot {g^\prime }(3) + {f_y}(2,7) \cdot {h^\prime }(3)\\ = 6 \cdot 5 + ( - 8) \cdot ( - 4)\\ = 30 + 32\end{aligned}\)

That is, \({\left. {\frac{{\partial z}}{{\partial t}}} \right|_{t = 3}} = 62.\)

Therefore, the derivative of \(\frac{{dz}}{{dt}}\) at \(t = 3\)is \(62.\)

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Essential Calculus: Early Transcendentals

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Short Answer

Determine the derivative \(\frac{{dz}}{{dt}}\)at the given value of \(t.\)The functions are \(z = f(x,y),x = g(t)\) and \({\rm{ }}y = h(t){\rm{.}}\)

Step by Step Solution

Step 1: The chain rule.

Step 2: Use the chain rule for calculation.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Determine the derivative \(\frac{{dz}}{{dt}}\)at the given value of \(t.\)The functions are \(z = f(x,y),x = g(t)\) and \({\rm{ }}y = h(t){\rm{.}}\)

Step by Step Solution

Step 1: The chain rule.

Step 2: Use the chain rule for calculation.

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