Use implicit differentiation to find the equation of the tangent line to the curve xy3+xy=14 at the point (7,1) . The equation of this tangent line can be written in the form y=mx+b
i don't seem to no how to find m or b
Take the derivative of both sides of the equation implicitly:
d/dx(xy^3+xy)=d/dx(14)
y^3+3xy^2*y'+y+xy'=0
At x=7, y=1. Thus, 1^3+3(7)(1^2)*y'+1+7y'=0
which yields y'=-2/28=-1/14
The slope of the tangent line is m which also is y'. Thus, you now have m=-1/14. Since the point (7,1) is on the tangent line, then just plug these coordinates as well as the value of m into the equation y=mx+b and solve for b.
Well, don't worry, let me clown around with this problem and help you out!
To find the equation of the tangent line to the curve, we need to use implicit differentiation. So, let's differentiate both sides of the equation with respect to x.
For the left side of the equation, using the product rule, the derivative of xy^3 is y^3 + 3xy^2 (dy/dx), and the derivative of xy is y + x(dy/dx).
On the right side of the equation, the derivative of 14 with respect to x is 0 since it's a constant.
So, putting all the pieces together, we have the derivative of xy^3 + xy equals 0, which simplifies to:
y^3 + 3xy^2 (dy/dx) + y + x(dy/dx) = 0.
Now, to find the slope of the tangent line (m), we can substitute the x and y values of the point (7,1) into the equation and solve for dy/dx.
Plugging in x = 7 and y = 1, we get:
1^3 + 3(7)(1)^2 (dy/dx) + 1 + 7(dy/dx) = 0.
1 + 21(dy/dx) + 1 + 7(dy/dx) = 0.
Simplifying, we have:
28(dy/dx) = -2.
Now, divide both sides by 28 to solve for dy/dx:
(dy/dx) = -2/28.
Reducing it further, dy/dx equals -1/14.
Great! We've found the slope of the tangent line (m). Now let's find the y-intercept (b).
We can substitute the coordinates of the point (7,1) and the value of m into the equation y = mx + b and solve for b. Using m = -1/14, x = 7, and y = 1, we have:
1 = (-1/14)(7) + b.
Simplifying, we get:
1 = -1/2 + b.
Rearranging the equation, we find:
b = 1 + 1/2.
Adding the fractions, we get b = 3/2.
So, the equation of the tangent line can be written in the form y = mx + b, where m = -1/14 and b = 3/2.
Putting it all together, the equation of the tangent line is:
y = (-1/14)x + 3/2.
Voilà! We've found the slope and intercept of the tangent line using implicit differentiation.
To find the equation of the tangent line at the point (7,1) on the curve xy^3+xy=14, we will use implicit differentiation to find the derivative dy/dx.
Let's start by differentiating both sides of the equation with respect to x:
d/dx(xy^3 + xy) = d/dx(14)
Using the product rule and the chain rule, the derivative of xy^3 + xy with respect to x is:
y^3 * dx/dx + 3xy^2 * dy/dx + y * dx/dx + x * dy/dx = 0
Simplifying this expression, we get:
1 + 3xy^2 * dy/dx + y + x * dy/dx = 0
Now, let's substitute the point (7,1) into the equation to find dy/dx:
1 + 3(7)(1)^2 * dy/dx + 1 + 7 * dy/dx = 0
Simplifying this expression, we get:
1 + 21 * dy/dx + 1 + 7 * dy/dx = 0
Combining like terms, we have:
28 * dy/dx = -2
Solving for dy/dx, we get:
dy/dx = -2/28 = -1/14
Now we have the slope of the tangent line at the point (7,1), which is m = -1/14. To find the y-intercept (b) of the tangent line, we substitute the coordinates (7,1) and the slope (-1/14) into the equation y = mx + b and solve for b:
1 = (-1/14)(7) + b
Simplifying this equation, we get:
1 = -7/14 + b
1 = -1/2 + b
Combining like terms, we have:
1 + 1/2 = b
3/2 = b
So the y-intercept (b) of the tangent line is 3/2.
Therefore, the equation of the tangent line to the curve xy^3 + xy = 14 at the point (7,1) can be written in the form y = mx + b as:
y = (-1/14)x + 3/2
To find the equation of the tangent line to the curve at the point (7,1), we can use implicit differentiation. Implicit differentiation is a technique used to differentiate equations where both x and y appear.
Let's differentiate the given equation with respect to x.
Start by differentiating each term of the equation using the product rule and treating y as a function of x.
The given equation is: xy^3 + xy = 14
Differentiating both sides with respect to x, we get:
(d/dx)(xy^3) + (d/dx)(xy) = (d/dx)(14)
To differentiate xy^3, we apply the product rule:
(d/dx)(xy^3) = x * (d/dx)(y^3) + y^3 * (d/dx)(x)
Since y is treated as a function of x, (d/dx)(y^3) can be written as (d/dx)(y^3) * (dy/dx). Similarly, (d/dx)(x) can be written as 1 * (dx/dx) = 1.
So, (d/dx)(xy^3) = x * 3y^2 * (dy/dx) + y^3 * 1 = 3xy^2 * (dy/dx) + y^3
Similarly, (d/dx)(xy) can be written as (d/dx)(xy) * (dy/dx):
(d/dx)(xy) = x * (d/dx)(y) + y * (d/dx)(x) = x * (dy/dx) + y * 1 = x * (dy/dx) + y
Now, let's substitute these values back into the differentiated equation:
3xy^2 * (dy/dx) + y^3 + x * (dy/dx) + y = 0
Now, we need to find the value of dy/dx at the point (7,1), which is the slope of the tangent line at that point.
To do this, we substitute x = 7 and y = 1 into the differentiated equation:
3(7)(1)^2 * (dy/dx) + (1)^3 + 7 * (dy/dx) + 1 = 0
Simplifying this equation, we get:
21(dy/dx) + 1 + 7(dy/dx) + 1 = 0
28(dy/dx) + 2 = 0
28(dy/dx) = -2
(dy/dx) = -2/28
(dy/dx) = -1/14
Therefore, the slope of the tangent line at the point (7,1) is -1/14.
Now, we can use the point-slope formula to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values of the point (7,1) and the slope -1/14, we have:
y - 1 = (-1/14)(x - 7)
Simplifying this equation, we get:
y - 1 = (-1/14)x + 1
Adding 1 to both sides, we have:
y = (-1/14)x + 2
Thus, the equation of the tangent line to the curve at the point (7,1) is y = (-1/14)x + 2, which is in the form y = mx + b with m = -1/14 and b = 2.