Calculate y' and y'' at the point (1,1) on the curve 5 xy^2+9 y-14 = 0
I got the first part (-5/19) but I can't get the answer for the second
first derivative:
5x y^2 + 9y - 14 = 0
5x(2y)dy/dx + 5y^2 + 9dy/dx = 0
dy/dx(10xy + 9) = -5y^2
dy/dx = -5y^2/(10xy+9)
at (1,1)
dy/dx
= y'
= -5/19 , you had that
from 5x(2y)dy/dx + 5y^2 + 9dy/dx = 0
10xy y' + 5y^2 + 9 y' = 0
recall the rule for derivatives of triple products
d(uvw) = uvw' + uv' w + u' vw
10(1)(y)(y') + 10x(y')(y') + 10xy(y'') + 10y y' + 9 y'' = 0
we can replace y' with -5/19
so .....
10(1)(1)(-5/19) + 10(1)(-5/19)(-5/19) + 10(1)(1)y'' + 10(1)(-5/19) + 9 y'' = 0
I will leave the arithmetic up to you
or, differentiating y' directly,
y" = 50y^3(-5xy^2+20xy+18)/(10xy+9)^3
y"(1,1) = 50(-5+20+18)/19^3 = 1650/6859
which agrees with Reiny's result above!!!
To find y' and y'' at the point (1,1) on the curve, we first need to differentiate the given equation implicitly with respect to x.
The given equation is:
5xy^2 + 9y - 14 = 0
Differentiating both sides of the equation with respect to x, using the product rule, we get:
(5y^2 + 10xy) * dy/dx + 9dy/dx = 0
Simplifying the equation, we get:
(5y^2 + 10xy + 9) * dy/dx = 0
Now, we can substitute the coordinates of the point (1,1) into the equation.
At (1,1), we have y = 1:
(5(1)^2 + 10(1)(1) + 9) * dy/dx = 0
(5 + 10 + 9) * dy/dx = 0
24 * dy/dx = 0
Since the derivative of a constant is 0, we find that dy/dx = 0.
Therefore, y' (the derivative of y with respect to x) at the point (1,1) is 0.
To find y'' (the second derivative of y with respect to x), we need to differentiate the equation obtained earlier, which was:
(5y^2 + 10xy + 9) * dy/dx = 0
Differentiating both sides with respect to x, using the product rule again, we have:
(10y + 10xy') * dy/dx + (5y^2 + 10xy + 9) * d^2y/dx^2 = 0
Substituting y = 1 and dy/dx = 0 (since we found y' to be 0 at (1,1)), we get:
(10(1) + 10(1)(0)) * 0 + (5(1)^2 + 10(1)(1) + 9) * d^2y/dx^2 = 0
(10 + 0) * 0 + (5 + 10 + 9) * d^2y/dx^2 = 0
24 * d^2y/dx^2 = 0
Since the derivative of a constant is 0, we find that d^2y/dx^2 = 0.
Therefore, y'' (the second derivative of y with respect to x) at the point (1,1) is 0.
To find the second derivative of a function, you will need to find the first derivative first. Let's start by finding the first derivative, y'.
Given the equation 5xy^2 + 9y - 14 = 0, we can differentiate implicitly with respect to x:
Differentiating each term with respect to x, we have:
(5xy^2)' + (9y)' - (14)' = 0
To differentiate (5xy^2) with respect to x, we use the Product Rule:
(5xy^2)' = 5(xy^2)' + (5x)'y^2
= 5(y^2 + 2xyy') + 5xy^2
Next, differentiating (9y) with respect to x gives:
(9y)' = 9y'
Since (14)' equals 0 because it is a constant term, we can rearrange the equation to solve for y':
5(y^2 + 2xyy') + 5xy^2 + 9y' = 0
Simplifying further, we have:
5y^2 + 10xyy' + 5xy^2 + 9y' = 0
Combining like terms, we get:
10xyy' + 5y^2 + 5xy^2 + 9y' = 0
Rearranging the terms, we obtain:
(10xyy' + 5xy^2) + (9y' + 5y^2) = 0
Factoring out common terms, we have:
5xy(y' + y) + 9(y' + y^2) = 0
Now, we can solve for y':
5xy(y' + y) = -9(y' + y^2)
Dividing both sides by (y' + y), we get:
5xy = -9(y' + y^2) / (y' + y)
Next, substitute the point (1,1) into the equation to solve for y':
5(1)(1)(y' + 1) = -9(y' + 1^2) / (y' + 1)
5(y' + 1) = -9(y' + 1)
5y' + 5 = -9y' - 9
Combining like terms, we get:
14y' = -14
Finally, dividing both sides by 14, we find the value of y':
y' = -1
Now let's find the second derivative y''.
To find y'', differentiate y' with respect to x:
Taking the derivative of y' = -1, we get:
(y')' = 0
Therefore, the second derivative y'' is 0.
Hence, at the point (1,1) on the curve, the first derivative y' is -1, and the second derivative y'' is 0.