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### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Find the partial derivative $$\frac{{\partial z}}{{\partial s}}$$and $$\frac{{\partial z}}{{\partial t}}$$ with the help of chain rule. The functions are $$z = {x^2}{y^3},x = s\cos t$$ and $$y = s\sin t.$$

The partial derivative $$\frac{{\partial z}}{{\partial s}}$$ and $$\frac{{\partial z}}{{\partial t}}$$ with the help of chain rule are $$2x{y^3}\cos t + 3{x^2}{y^2}\sin t$$ and $$- 2sx{y^3}\sin t + 3s{x^2}{y^2}\cos t$$ respectively.

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.

Differentiate $$z$$ partially with respect to $$s$$ as follows.

$$\frac{{\partial z}}{{\partial s}} = \frac{{\partial z}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial s}} + \frac{{\partial z}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial s}}$$

Differentiate $$z$$ partially with respect to $$x$$ as follows.

\begin{aligned}{l}\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{x^2}{y^3}} \right)\\ = {y^3}\frac{\partial }{{\partial x}}\left( {{x^2}} \right)\\ = {y^3}(2x)\\ = 2x{y^3}\end{aligned}

Differentiate $$z$$ partially with respect to $$y$$ as follows.

\begin{aligned}{l}\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{x^2}{y^3}} \right)\\ = {x^2}\frac{\partial }{{\partial y}}\left( {{y^3}} \right)\\ = {x^2}\left( {3{y^2}} \right)\\ = 3{x^2}{y^2}\end{aligned}

Differentiate $$x$$ partially with respect to $$s$$ as follows.

$$\frac{{\partial x}}{{\partial s}} = \frac{\partial }{{\partial s}}(s\cos t) = \cos t$$

Differentiate $$y$$ partially with respect to $$s$$ as follows.

$$\frac{{\partial y}}{{\partial s}} = \frac{\partial }{{\partial s}}(s\sin t) = \sin t$$

Then, the partial derivative of$$z$$ with respect to$$s$$ becomes.

\begin{aligned}{l}\frac{{\partial z}}{{\partial s}} = \left( {2x{y^3}} \right)(\cos t) + \left( {3{x^2}{y^2}} \right)(\sin t)\\ = 2x{y^3}\cos t + 3{x^2}{y^2}\end{aligned}