Find the first partial derivatives of the function. z=xsin⁡(xy)

Question

Macsen Nixon · Accepted Answer

f(x,y)=xsin⁡(xy)  To find fx we trat y as constant and differentiate withe respect to x  fx=sin⁡(xy)+xycos⁡(xy)  To find fy we treat x as constant and differentiate with respect to y  fy=x2cos⁡(xy)  Result  fx(x,y)=sin⁡(xy)+xycos⁡(xy)  fy=x2cos⁡(xy)

Don Sumner · Accepted Answer

Answer:x2cos(xy)Explanation:First, let&#039;s find the partial derivative with respect to x, denoted as ∂z∂x. To differentiate z with respect to x, we treat y as a constant and apply the chain rule. The chain rule states that if we have a function f(g(x)), then its derivative is given by dfdx=dfdg·dgdx.Using the chain rule, we differentiate z with respect to x as follows:∂z∂x=∂∂x(xsin(xy))=sin(xy)+xcos(xy)·∂∂x(xy)Since ∂∂x(xy) is simply y, we can simplify the expression:∂z∂x=sin(xy)+xycos(xy)Now, let&#039;s find the partial derivative with respect to y, denoted as ∂z∂y. Again, we treat x as a constant and differentiate z with respect to y using the chain rule:∂z∂y=∂∂y(xsin(xy))=xcos(xy)·∂∂y(xy)Since ∂∂y(xy) is simply x, we have:∂z∂y=xcos(xy)·x=x2cos(xy)Therefore, the first partial derivatives of the function z=xsin(xy) are:∂z∂x=sin(xy)+xycos(xy)∂z∂y=x2cos(xy)

RizerMix · Accepted Answer

To find the first partial derivatives of the function

z = x \sin (x y)

, we need to differentiate with respect to

x

and

y

separately.
The partial derivative with respect to

x

(

\frac{\partial z}{\partial x}

) is obtained by treating

y

as a constant and differentiating

x \sin (x y)

with respect to

x

:

\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (x \sin (x y)) = \sin (x y) + x \cos (x y) \cdot y = \sin (x y) + x y \cos (x y) .

The partial derivative with respect to

y

(

\frac{\partial z}{\partial y}

) is obtained by treating

x

as a constant and differentiating

x \sin (x y)

with respect to

y

:

\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (x \sin (x y)) = x \cos (x y) \cdot x = x^{2} \cos (x y) .

nick1337 · Accepted Answer

Step 1: Let&#039;s begin by finding the partial derivative with respect to x, denoted as ∂z∂x. To differentiate sin(xy) with respect to x, we treat y as a constant:∂z∂x=∂∂x(xsin(xy))Using the product rule, the derivative of x with respect to x is 1, and the derivative of sin(xy) with respect to x is cos(xy) multiplied by the derivative of xy with respect to x, which is simply y:∂z∂x=1·sin(xy)+x·cos(xy)·ySimplifying further, we have:∂z∂x=sin(xy)+xycos(xy)Step 2: Now, let&#039;s find the partial derivative with respect to y, denoted as ∂z∂y. Again, we treat x as a constant:∂z∂y=∂∂y(xsin(xy))Using the product rule, the derivative of x with respect to y is 0 (since x is a constant), and the derivative of sin(xy) with respect to y is cos(xy) multiplied by the derivative of xy with respect to y, which is simply x:∂z∂y=0·sin(xy)+x·cos(xy)·xSimplifying further, we have:∂z∂y=x2cos(xy)Step 3:Therefore, the first partial derivatives of the function z=xsin(xy) are:∂z∂x=sin(xy)+xycos(xy)and∂z∂y=x2cos(xy)

Find the first partial derivatives of the function. z = xsin(xy)

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