Find the slope of the tangent line to the curve (sqrt(4x+y))+(sqrt(3xy)) at the point (3,4)
The slope of the tangent line to the curve at the given point is ?
I got
-((1/2)+4sqrt(36))/((1/8)+3sqrt(36))
√(4x+y)+√(3xy)
is not a curve. It is a surface.
I guess you meant
√(4x+y)+√(3xy) = 10
Hmmm. Maybe not, since that does not lead to the value you give. Try again with the post, and maybe show some intermediate steps.
yeah, sorry the equation was √(4x+y)+√(3xy) = 10.
To find the slope of the tangent line to the curve at the point (3,4), we need to take the derivative of the curve with respect to x and evaluate it at x=3.
Let's find the derivative step by step:
1. The given curve is (sqrt(4x+y)) + (sqrt(3xy)). Let's call this function f(x, y).
2. To find df/dx, we need to use the chain rule and differentiate each term separately.
- Differentiating the first term, sqrt(4x+y), with respect to x:
d/dx [sqrt(4x+y)] = (1/2)*(4x+y)^(-1/2) * (4 + dy/dx)
- We use the chain rule here by multiplying the derivative of the inner function (4x+y) with respect to x, by the derivative of the outer function (sqrt(u)) which is (1/2)u^(-1/2).
- Differentiating the second term, sqrt(3xy), with respect to x:
d/dx [sqrt(3xy)] = (1/2)*(3xy)^(-1/2) * (3y + x(dy/dx))
3. Combining the two derivatives, we have:
df/dx = (1/2)*(4x+y)^(-1/2) * (4 + dy/dx) + (1/2)*(3xy)^(-1/2) * (3y + x(dy/dx))
4. At the point (3,4), we substitute x=3, y=4 into the equation we obtained in the previous step to find the slope of the tangent line:
slope = (1/2)*(4(3)+4)^(-1/2) * (4 + dy/dx) + (1/2)*(3(3)(4))^(-1/2) * (3(4) + 3(dy/dx))
= (1/2)*(16)^(-1/2) * (4 + dy/dx) + (1/2)*(36)^(-1/2) * (12 + 3(dy/dx))
= (1/2)*(1/4) * (4 + dy/dx) + (1/2)*(1/6) * (12 + 3(dy/dx))
= (1/8) * (4 + dy/dx) + (1/12) * (12 + 3(dy/dx))
= (1/8)*(4) + (1/8)*(dy/dx) + (1/12)*(12) + (1/12)*(3(dy/dx))
= 1/2 + (1/8)*(dy/dx) + 1 + (1/4)*(dy/dx)
5. Simplifying the expression:
slope = 1/2 + (1/8)*(dy/dx) + 1 + (1/4)*(dy/dx)
= 3/2 + (3/8)*(dy/dx)
Therefore, the slope of the tangent line to the curve at the point (3,4) is 3/2 + (3/8)*(dy/dx).
To find the slope of the tangent line to a curve at a given point, we can use the concept of derivatives. The derivative represents the rate at which the function is changing at a specific point.
To start, let's find the derivative of the function f(x) = √(4x + y) + √(3xy):
Step 1: Use the power rule to differentiate the first term, √(4x + y).
The derivative of √(4x + y) with respect to x is (1/2) * (4x + y)^(-1/2) * (4).
Step 2: Use the chain rule to differentiate the second term, √(3xy).
The derivative of √(3xy) with respect to x is (1/2) * (3xy)^(-1/2) * 3y.
Step 3: Combine the derivatives of the two terms:
The derivative of f(x) is (1/2) * (4x + y)^(-1/2) * 4 + (1/2) * (3xy)^(-1/2) * 3y.
Now that we have the derivative, we can find the slope of the tangent line at a specific point by substituting the x-coordinate and y-coordinate of that point into the derivative.
In this case, the given point is (3, 4). Substituting x = 3 and y = 4 into the derivative, we get:
[(1/2) * (4(3) + 4)^(-1/2) * 4] + [(1/2) * (3(3)(4))^(-1/2) * 3(4)]
= [(1/2) * (12 + 4)^(-1/2) * 4] + [(1/2) * (36)^(-1/2) * 3(4)]
= [(1/2) * (16)^(-1/2) * 4] + [(1/2) * (36)^(-1/2) * 12]
= (1/2) * 4/(√16) + (1/2) * 12/(√36)
= (1/2) * 2 + (1/2) * 6
= 1 + 3
= 4.
Therefore, the slope of the tangent line to the curve at the point (3, 4) is 4.