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tangare [24]
3 years ago
5

Calculate dy/dx when xy^3 + y =3x

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
6 0
You have to use implicit derivative

dy / dx = y'

xy^3+y=3x

y^3 + 3xy^2 y' + y' = 3

[3xy^2 + 1) y'= 3 - y^3

          3 - y^3
y' = -----------------   <-----------answer
         3xy^2 + 1

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6 0
3 years ago
Help please!!!!!!!!!
In-s [12.5K]

ANSWER

24


EXPLANATION

For a matrix A of order n×n, the cofactor C_{ij} of element a_{ij} is defined to be


   C_{ij} = (-1)^{i+j} M_{ij}


M_{ij} is the minor of element a_{ij} equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.


Here, we have


   C_{11} = (-1)^{1+1} M_{11} = M_{11}


M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.


   \begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right)  \end{aligned}


Since the determinant of a 2×2 matrix is


   \det\left(  \begin{bmatrix} a & b \\ c& d  \end{bmatrix} \right) = ad-bc


it follows that


   \begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}


so C_{11} = M_{11} = 24

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