if 3x^2+5x+xy=1 y(1)=-7 and y'(1)= ?
I can solve for dy/dx but I don't know what to do with y(1) and y'(1)
To find y'(1), you need to first find the derivative of y with respect to x (dy/dx). Then, substitute x = 1 into the derivative expression to evaluate y'(1).
Given the equation 3x^2 + 5x + xy = 1, it is a nonlinear equation that involves both x and y. To find dy/dx, we need to differentiate both sides of the equation with respect to x. This involves applying the product rule and the chain rule for differentiation.
Let's go through the steps to find y'(x):
1. Differentiate both sides of the equation with respect to x:
d/dx(3x^2 + 5x + xy) = d/dx(1)
2. Apply the product rule:
For the term 3x^2, we have: d/dx(3x^2) = 6x.
For the term 5x, we have: d/dx(5x) = 5.
For the term xy, we have: d/dx(xy) = x * (dy/dx) + y * (d/dx(x)).
Applying the product rule, we get: d/dx(xy) = x * (dy/dx) + y.
The right side of the equation remains unchanged since it's a constant, so its derivative is zero (d/dx(1) = 0).
Combining these results, we have: 6x + 5 + x * (dy/dx) + y = 0.
3. Rearrange the equation to solve for dy/dx:
x * (dy/dx) = -6x - 5 - y.
4. Divide both sides by x to isolate dy/dx:
dy/dx = (-6x - 5 - y) / x.
Now that we have the expression for dy/dx, we can substitute x = 1 to find y'(1):
y'(1) = (-6(1) - 5 - y) / 1.
Given that y(1) = -7, we can substitute this value in:
y'(1) = (-6(1) - 5 - (-7)) / 1.
Simplifying further:
y'(1) = (-6 - 5 + 7) / 1
= -4.
Therefore, y'(1) = -4.
x y = -3 x^2 - 5 x + 1 ... y = - 3 x - 5 + 1/x ... so, y(1) = - 3 - 5 + 1 = -7
y' = dy/dx = - 3 - 1/x^2 ... so, y'(1) = - 3 - 1 = -4