Let z=( f(x, y))n. Show that ∂z∂x=n( f(x, y))n-1∂f∂x and ∂z∂y=n( f(x, y))n-1∂f∂y .

∂z∂x=n( f(x, y))n-1∂f∂x∂z∂y=n( f(x, y))n-1∂f∂y

Q. 71 - Calculus | StudyQuestionHub

Q. 71

Question

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

Step-by-Step Solution

Verified

Answer

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

1Step 1. Given information

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

2Step 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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