find y' using the implicit differentiation : -x +3x^2 = y^2 -3x^5*y^2

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justaguide | College Teacher | (Level 2) Distinguished Educator

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Using implicit differentiation for -x +3x^2 = y^2 -3x^5*y^2, we get:

-x +3x^2 = y^2 -3x^5*y^2

using the product rule for 3x^5*y^2

=> -1 + 6x = 2y y' - 3x^5 * 2y y' - 15x^4* y^2

=> y'(2y - 6x^5*y ) = -1 + 6x + 15x^4*y^2

=> y' = -1 + 6x + 15x^4*y^2/ (2y - 6x^5*y )

Therefore the required result is (-1 + 6x + 15x^4*y^2)/ (2y - 6x^5*y )

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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Given the functions:

-x + 3x^2 = y^2 - 3x^5 y^2

We will use the implicit differentiation to fin y'.

(-x)' + 3(x^2)' = (y62)' -3(x^5*y^2)'

(-x)' + 3(x^2)' = (y^2)' -3[(x^5)'*y^2 + (x^5*(y^2)']

-1 + 6x = 2yy' - 3[ 5x^4*y^2 + x^5* 2yy']

-1 + 6x = 2yy' -15x^4*y^2 - 6x^5 *yy'

Now we will combine the terms with y' on the left sides.

==> 6x^5*yy' - 2yy' = 1-6x - 15x^4*y^2

Now we will factor y'.

==> y'( 6x^5 *y - 2y) = (1-6x -15x^4*y^2)

Now we will divide by (6x^5*y - 2y)

==> y' = ( 1- 6x-15x^4*y^2) / (6x^5 y - 2y)

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