Answer to Question #275498 in Calculus for Angel Nodado

Question

Find the first partial derivative of

f(w, x, y, z) = w ^ 2 * x * y ^ 2 + x * y ^ 3 * z ^ 2

Answer 1

$\frac{\partial f}{\partial x}(x,y,z,w)=\frac{\partial (x\cdot y^2\cdot w^2+x\cdot y^3\cdot z^2)}{\partial x}=\\$

$\frac{\partial (x\cdot y^2\cdot w^2)}{\partial x}+ \frac{\partial (x\cdot y^3\cdot z^2)}{\partial x}=y^2\cdot w^2\cdot \frac{\partial x}{\partial x}+y^3\cdot z^2\cdot \frac{\partial x}{\partial x}=y^2\cdot w^2\cdot 1+y^3\cdot z^2\cdot 1=\\ =y^2\cdot w^2+y^3\cdot z^2$

$\frac{\partial f}{\partial y}(x,y,z,w)=\frac{\partial (x\cdot y^2\cdot w^2+x\cdot y^3\cdot z^2)}{\partial y}=\\$

$=\frac{\partial (x\cdot y^2\cdot w^2)}{\partial y}+ \frac{\partial (x\cdot y^3\cdot z^2)}{\partial y}=x\cdot w^2\cdot \frac{\partial y^2}{\partial x}+x\cdot z^2\cdot \frac{\partial y^3}{\partial y}=x\cdot w^2\cdot (2y)+x\cdot z^2\cdot (3y^2)=\\ =2\cdot x\cdot y \cdot w^2+3\cdot x\cdot y^2\cdot z^2$

$\frac{\partial f}{\partial z}(x,y,z,w)=\frac{\partial (x\cdot y^2\cdot w^2+x\cdot y^3\cdot z^2)}{\partial z}=\\$

$\frac{\partial (x\cdot y^2\cdot w^2)}{\partial z}+ \frac{\partial (x\cdot y^3\cdot z^2)}{\partial z}=0+x\cdot y^3\cdot \frac{\partial z^2}{\partial z}=x\cdot y^3\cdot(2z)=2\cdot x\cdot y^3\cdot z$

$\frac{\partial f}{\partial w}(x,y,z,w)=\frac{\partial (x\cdot y^2\cdot w^2+x\cdot y^3\cdot z^2)}{\partial w}=\\$

$\frac{\partial (x\cdot y^2\cdot w^2)}{\partial w}+ \frac{\partial (x\cdot y^3\cdot z^2)}{\partial w}=x\cdot y^2\cdot \frac{\partial w^2}{\partial w}+0=x\cdot y^2\cdot (2w)=2\cdot x\cdot y^2\cdot w$