Respuesta :
x² + 2xy − y² + x = 6
When you implicitly differentiate, you differentiate the entire expression. Differentiate x terms normally, but use the chain rule when you have a term with y in it. Remember to use the chain and product rule on 2xy (think of 2xy as 2x*y).
2x + 2y + 2x*dy/dx - 2y*dy/dx + 1 = 0
Now separate dy/dx from everything else:
dy/dx*(2x - 2y) = -(2x + 2y + 1)
dy/dx = -(2x + 2y + 1)/(2x - 2y)
The slope is dy/dx, so this means at the point (2, 4), the slope is:
dy/dx = -(2x + 2y + 1)/(2x - 2y)
dy/dx = -(2(2) + 2(4) + 1)/(2(2) - 2(4))
= -(4 + 8 + 1)/(4 - 8)
= 13/4
Using point slope form, the tangent line is:
y - y1 = m(x - x1)
y - 4 = 13/4*(x - 2)
When you implicitly differentiate, you differentiate the entire expression. Differentiate x terms normally, but use the chain rule when you have a term with y in it. Remember to use the chain and product rule on 2xy (think of 2xy as 2x*y).
2x + 2y + 2x*dy/dx - 2y*dy/dx + 1 = 0
Now separate dy/dx from everything else:
dy/dx*(2x - 2y) = -(2x + 2y + 1)
dy/dx = -(2x + 2y + 1)/(2x - 2y)
The slope is dy/dx, so this means at the point (2, 4), the slope is:
dy/dx = -(2x + 2y + 1)/(2x - 2y)
dy/dx = -(2(2) + 2(4) + 1)/(2(2) - 2(4))
= -(4 + 8 + 1)/(4 - 8)
= 13/4
Using point slope form, the tangent line is:
y - y1 = m(x - x1)
y - 4 = 13/4*(x - 2)
