Question textLet [math: f(x,y)=ysin⁡(xy)]f(x,y) = {y\,\sin \left( x\,y \right)}. The partial derivative [math: ∂f∂x]\frac{\partial f}{\partial {x}} is[input] Your answer should contain the variables [math: x] and [math: y].Check Question 66

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BlackTom AI · Accepted Answer

Let's break down the function step by step and apply differentiation rules carefully.
The function is f(x,y) = y * sin(xy). Here, when differentiating with respect to x, treat y as a constant multiplier (since partial derivative with respect to x holds y fixed).
Next, apply the chain rule to the sin(xy) part. If we let u = x y, then sin(u) has derivative cos(u) * du/dx. Since u = x y, du/dx = y (because derivative of x with respect to x is 1 and y is treated as a constant multiplier).
Putting these together: d/dx [sin(xy)] = cos(xy) * y.
Now multiply by the outer constant factor y (the other part of f): d/dx [y * sin(xy)] = y * [cos(xy) * y] = y^2 * cos(xy).
Thus, the partial derivative ∂f/∂x is y^2 cos(xy).
A common confusion is to drop the chain rule for the inner argument xy or to forget the outer y factor; both would lead to incorrect results. For instance, forgetting that du/dx = y when u = xy would yield an error, and treating sin(xy) as if its derivative were cos(xy) times x’d only would miss the y factor entirely.
In summary, the correct result is y^2 cos(xy).

ENG1005 - MUM S2 2025 [FINAL REVISION] Quizzes

Question textLet [math: f(x,y)=ysin⁡(xy)]f(x,y) = {y\,\sin \left( x\,y \right)}. The partial derivative [math: ∂f∂x]\frac{\partial f}{\partial {x}} is[input] Your answer should contain the variables [math: x] and [math: y].Check Question 66

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