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If z = sin (x^2 + y^2), x = u cos (v), y = u sin (v), find partial z / partial u and partial z /...

Question:

If {eq}\displaystyle z = \sin (x^2 + y^2),\ x = u \cos (v),\ y = u \sin (v) {/eq}, find {eq}\dfrac {\partial z}{\partial u} {/eq} and {eq}\dfrac {\partial z}{\partial v} {/eq}. The variables are restricted to domains on which the functions are defined.

Chain Rule of Derivatives:

Chain rule of derivative is used to find out the derivative of composite function i.e. function inside another function. It can be given by the following formula:

{eq}\frac {d}{dx} f(g(x) ) = f'(g(x)) g'(x) {/eq}

Power rule of derivative is used to find out the derivative having the form {eq}x^n. {/eq} It is given by the following formula:

{eq}\frac {d}{dx} x^n = nx^{n-1} {/eq}

Also, some useful formulae are the following:

{eq}\frac {d}{dx} \sin x = \cos x {/eq}

When the function that has to be differentiated consist of two independent variables, then it has to be differentiated partially i.e. differentiating with respect to one independent variable only, taking another independent variable as constant.

Answer and Explanation:

Given:

{eq}\displaystyle z = \sin (x^2 + y^2),\ x = u \cos (v),\ y = u \sin (v) \\ {/eq}

On putting the values of 'x' and 'y' into the expression of 'z', we get:

{eq}\displaystyle { z = \sin (x^2 + y^2) \\ = \sin ( ( u \cos (v))^2 + (u \sin (v))^2) \\ = \sin ( u^2 \cos^2 (v) + u^2 \sin^2 (v) ) \\ = \sin ( u^2 ( \cos^2 (v) + \sin^2 (v) ) ) \\ = \sin ( u^2 ( 1 ) ) \\ = \sin ( u^2 ) \\ } {/eq}

On differentiating the above expression partially with respect to 'u', we get:

{eq}\dfrac {\partial z}{\partial u} = \cos (u^2) \frac {\partial }{\partial u} (u^2) \\ = \cos (u^2) (2u) \\ {/eq}

which is the answer.

On differentiating the above expression partially with respect to 'v', we get:

{eq}\dfrac {\partial z}{\partial v} = 0 \\ {/eq}

which is the answer.


Learn more about this topic:

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The Chain Rule for Partial Derivatives

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 4
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