Chapter 12: Q. 71 (page 945)

$Let z = (f {(x, y))}^{n} . Show that \frac{\partial z}{\partial x} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial x} and \frac{\partial z}{\partial y} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial y} .$

Short Answer

Expert verified

$\frac{\partial z}{\partial x} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial x} \frac{\partial z}{\partial y} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial y}$

Step by step solution

Step 1. Given information

A function, $z = (f {(x, y))}^{n}$

Step 2. Proofs of given partial derivatives

$LHS = \frac{\partial z}{\partial x} = \frac{\partial (f {(x, y))}^{n}}{\partial x} Using power rule, chain rule and definition of partial derivatives, it follows that : = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial x} = RHS Similarly, for partial derivative with respect to y, we get, LHS = \frac{\partial z}{\partial y} = \frac{\partial (f {(x, y))}^{n}}{\partial y} Using power rule, chain rule and definition of partial derivatives, it follows that : = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial y} = RHS$

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$Let z = (f {(x, y))}^{n} . Show that \frac{\partial z}{\partial x} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial x} and \frac{\partial z}{\partial y} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial y} .$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proofs of given partial derivatives

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Letz=(f(x,y))n.Showthat∂z∂x=n(f(x,y))n-1∂f∂xand∂z∂y=n(f(x,y))n-1∂f∂y.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Proofs of given partial derivatives

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

$Let z = (f {(x, y))}^{n} . Show that \frac{\partial z}{\partial x} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial x} and \frac{\partial z}{\partial y} = n (f {(x, y))}^{n - 1} \frac{\partial f}{\partial y} .$