Suppose z = xy + tan^{-1}(x + y^{2}) x = st^{2} +1, y=te^{s} Find \frac{\partial z}{\partial s}...


Suppose {eq}z = xy + tan^{-1}(x + y^{2}){/eq}

{eq}x = st^{2} +1{/eq}, {eq}null{/eq}

Find {eq}\frac{\partial z}{\partial s}{/eq} when {eq}s = 0{/eq} and {eq}t = 2{/eq}

Partial derivative:

A Partial derivative of a function of a several variables is its derivative with respect to one of those variables,

with the other variables held constant.


{eq}\frac{\partial }{{\partial x}}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{1 + {x^2}}} {/eq}

Answer and Explanation:

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Given function of several variable

{eq}z = xy + {\tan ^{ - 1}}\left( {x + {y^2}} \right) {/eq}

putting {eq}x = s{t^2} + 1,y = t{e^5} {/eq}, in z...

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Learn more about this topic:

Partial Derivative: Definition, Rules & Examples


Chapter 18 / Lesson 12

When a function depends on more than one variable, we can use the partial derivative to determine how that function changes with respect to one variable at a time. In this lesson, we use examples to define partial derivatives and to explain the rules for evaluating them.

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