00:02
To find an equation of the tangent line at the curve at 5 .1, we need to find the slope of the tangent line.
00:07
And to find the slope of the tangent line, we can use the formula.
00:10
Slope is equal to df, d, f, d, d, f, d, d .x, over df, dx.
00:23
And f is the function that describes the curve, and df, d, d, y, are the partial derivatives of the function with respect to x and y respectively.
00:29
So given the original equation 2xy cubed plus xy equals 15, the partial derivatives of the function with respect to x would be 2 y cubed plus y and with respect to y would be 6 y squared or excuse me 6x y squared plus x and plugging the values that are given in an equation, we would get a slope of df, d, f, d, y, which is going to be six times five times, and i'm substituting here, x, y from 5, 1.
01:21
So, df, d, y, in the numerator, 6 times 5 times 1 squared, plus 5 over in the denominator, df, dx, 2, y, q plus y...
