Check the solution to "Find both first partial derivatives. f(x,y)=x2−2y2+4" - Plainmath

Alyce Wilkinson

Answered question

2020-11-23

Find both first partial derivatives.

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

Answer & Explanation

Leonard Stokes

Skilled2020-11-24Added 98 answers

Step 1
we have to find the first partial derivatives:
the given function is:

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

Step 2

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

= \frac{\partial x^{2}}{\partial x} - 2 \frac{\partial x^{2}}{\partial x} + \frac{\partial 4}{\partial x}

= 2 x - 2 (0) + 0

= 2 x - 0 + 0 = 2 x

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

= \frac{\partial x^{2}}{\partial y} - \frac{\partial 2 y^{2}}{\partial y} + \frac{\partial 4}{\partial y}

= 0 - 2 \frac{\partial y^{2}}{\partial y} + 0

= - 2 (2 y) = - 4 y

Step 3
therefore the value of both first partial derivatives is:

\frac{\partial f}{\partial x} = 2 x

\frac{\partial f}{\partial y} = - 4 y

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