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Expert-verified Found in: Page 656 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # 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 $$62.$$

See the step by step solution

## 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 $$62.$$ ### Want to see more solutions like these? 