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find $f_x(x,y,z)$, $f_y(x,y,z)$, $f_z(x,y,z)$, and $f_{yx}(x,y,z)$ for …

Question

find $f_x(x,y,z)$, $f_y(x,y,z)$, $f_z(x,y,z)$, and $f_{yx}(x,y,z)$ for the following $f(x,y,z)=4x^5 + 3xy - 3z^5$ $f_x(x,y,z)=$ $f_y(x,y,z)=$ $f_z(x,y,z)=$ $f_{yx}(x,y,z)=$

Explanation:

Step1: Find \(f_x(x,y,z)\)

Differentiate \(f(x,y,z) = 4x^5+3xy - 3z^5\) with respect to \(x\).
Using the power rule \(\frac{d}{dx}(x^n)=nx^{n - 1}\) and treating \(y\) and \(z\) as constants.
\(f_x(x,y,z)=\frac{\partial}{\partial x}(4x^5)+\frac{\partial}{\partial x}(3xy)-\frac{\partial}{\partial x}(3z^5)\)
\(f_x(x,y,z)=4\times5x^{4}+3y-0 = 20x^{4}+3y\)

Step2: Find \(f_y(x,y,z)\)

Differentiate \(f(x,y,z) = 4x^5+3xy - 3z^5\) with respect to \(y\).
Using the power rule and treating \(x\) and \(z\) as constants.
\(f_y(x,y,z)=\frac{\partial}{\partial y}(4x^5)+\frac{\partial}{\partial y}(3xy)-\frac{\partial}{\partial y}(3z^5)\)
\(f_y(x,y,z)=0 + 3x-0=3x\)

Step3: Find \(f_z(x,y,z)\)

Differentiate \(f(x,y,z) = 4x^5+3xy - 3z^5\) with respect to \(z\).
Using the power rule and treating \(x\) and \(y\) as constants.
\(f_z(x,y,z)=\frac{\partial}{\partial z}(4x^5)+\frac{\partial}{\partial z}(3xy)-\frac{\partial}{\partial z}(3z^5)\)
\(f_z(x,y,z)=0+0-3\times5z^{4}=- 15z^{4}\)

Step4: Find \(f_{yx}(x,y,z)\)

Differentiate \(f_y(x,y,z) = 3x\) with respect to \(x\).
Treating \(y\) and \(z\) as constants.
\(f_{yx}(x,y,z)=\frac{\partial}{\partial x}(3x)=3\)

Answer:

\(f_x(x,y,z)=20x^{4}+3y\), \(f_y(x,y,z) = 3x\), \(f_z(x,y,z)=-15z^{4}\), \(f_{yx}(x,y,z)=3\)