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Exercise. Compute the partial derivative of f(x,y)=x/yf(x, y)=x / y, $f(x, y, z)=x \sin...
Jun 13, 2024
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Solution by Steps
step 1
Compute the partial derivative of f(x,y)=xyf(x, y) = \frac{x}{y} with respect to xx
step 2
x(xy)=1y\frac{\partial}{\partial x} \left( \frac{x}{y} \right) = \frac{1}{y}
step 3
Compute the partial derivative of f(x,y)=xyf(x, y) = \frac{x}{y} with respect to yy
step 4
y(xy)=xy2\frac{\partial}{\partial y} \left( \frac{x}{y} \right) = -\frac{x}{y^2}
step 5
Compute the partial derivative of f(x,y,z)=xsin(yz)f(x, y, z) = x \sin(yz) with respect to xx
step 6
x(xsin(yz))=sin(yz)\frac{\partial}{\partial x} \left( x \sin(yz) \right) = \sin(yz)
step 7
Compute the partial derivative of f(x,y,z)=xsin(yz)f(x, y, z) = x \sin(yz) with respect to yy
step 8
y(xsin(yz))=xzcos(yz)\frac{\partial}{\partial y} \left( x \sin(yz) \right) = xz \cos(yz)
step 9
Compute the partial derivative of f(x,y,z)=xsin(yz)f(x, y, z) = x \sin(yz) with respect to zz
step 10
z(xsin(yz))=xycos(yz)\frac{\partial}{\partial z} \left( x \sin(yz) \right) = xy \cos(yz)
Answer
The partial derivatives are: 1. x(xy)=1y\frac{\partial}{\partial x} \left( \frac{x}{y} \right) = \frac{1}{y} 2. y(xy)=xy2\frac{\partial}{\partial y} \left( \frac{x}{y} \right) = -\frac{x}{y^2} 3. x(xsin(yz))=sin(yz)\frac{\partial}{\partial x} \left( x \sin(yz) \right) = \sin(yz) 4. y(xsin(yz))=xzcos(yz)\frac{\partial}{\partial y} \left( x \sin(yz) \right) = xz \cos(yz) 5. z(xsin(yz))=xycos(yz)\frac{\partial}{\partial z} \left( x \sin(yz) \right) = xy \cos(yz)
Key Concept
Partial Derivatives
Explanation
Partial derivatives measure how a function changes as its variables change. They are essential in multivariable calculus for understanding the behavior of functions with more than one variable.
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