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MATH1062/1005/1023 (ND) MATH1062/1023 Calculus Quiz 10

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Consider the function ๐‘“ ( ๐‘ฅ , ๐‘ฆ ) = sin โก ( 5 ๐‘ฅ ๐‘ฆ ) . What is ๐‘“ ๐‘ฆ ๐‘ฅ ?

Options
A.25 ๐‘ฆ 3 sin โก ( 5 ๐‘ฅ ๐‘ฆ )
B.โˆ’ 25 ๐‘ฅ ๐‘ฆ 4 sin โก ( 5 ๐‘ฅ ๐‘ฆ )
C.25 ๐‘ฅ ๐‘ฆ 3 sin โก ( 5 ๐‘ฅ ๐‘ฆ ) โˆ’ 25 ๐‘ฆ 2 cos โก ( 5 ๐‘ฅ ๐‘ฆ )
D.25 ๐‘ฆ 2 sin โก ( 5 ๐‘ฅ ๐‘ฆ ) โˆ’ 25 ๐‘ฅ ๐‘ฆ 3 cos โก ( 5 ๐‘ฅ ๐‘ฆ )
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We start by restating the problem: given f(x,y) = sin(5xy), determine the mixed partial derivative f_yx, i.e., the derivative first with respect to y and then with respect to x. Option 1: 25 y^3 sin(5xy). This expression involves only a sine factor with a cubic power of y and lacks any cosine term or a product with x; it does not match the structure of a derivative of sin(5xy) with respect to y and then x. The original function depends on xy inside the sine, so derivatives typically bring down cosine or sine factors multiplied by powers of x or y; this option is missing those essential components. Option 2: โˆ’25 x y^4 sin(5xy). Here we see a sine factor sin(5xy) multiplied by a p......Login to view full explanation

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