Key to "Find the first partial derivatives of f(x,y)=4xcos⁡(x2+y2)." - Plainmath

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Answered question

2021-09-18

Find the first partial derivatives of

f (x, y) = 4 x \cos (x^{2} + y^{2})

.

Answer & Explanation

Viktor Wiley

Skilled2021-09-19Added 84 answers

Step 1
Given

f (x, y) = 4 x \cos (x^{2} + y^{2})

Step 2

f (x, y) = 4 x \cos (x^{2} + y^{2})

First partial derivatives
Take y as constant

\frac{\partial f}{\partial x} = 4 \frac{\partial}{\partial x} (x \cos (x^{2} + y^{2}))

= 4 [x \frac{\partial}{\partial x} \cos (x^{2} + y^{2})] + 4 \cos (x^{2} + y^{2}) \frac{\partial x}{\partial x}

= - 4 x \sin (x^{2} + y^{2}) (2 x) + 4 \cos (x^{2} + y^{2})

= - 8 x^{2} \sin (x^{2} + y^{2}) + 4 \cos (x^{2} + y^{2})

take x as constant

\frac{\partial f}{\partial y} = 4 x \frac{\partial}{\partial y} (\cos (x^{2} + y^{2}))

= - 4 x \sin (x^{2} + y^{2}) (2 y)

= - 8 x y \sin (x^{2} + y^{2})

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