Use implicit differentiation to find the slope of the tangent line to the curve XY^3 + XY = 5 at the point (3,2) .
3Y^2 x dy+ Y^3 dx+ x dy+ y dx=0
dy(3xy^2+x)=-dx(y+y^3)
dy/dx= slope= -(y^3+y)/(3xy^2+x)
put in the x,y and solve.
check my work, I did it in a hurry.
i got the answer right thanks
To find the slope of the tangent line to a curve using implicit differentiation, follow these steps:
Step 1: Differentiate the equation implicitly with respect to x, treating y as a function of x.
Let's differentiate the equation XY^3 + XY = 5 with respect to x.
Differentiating both sides of the equation gives:
(d/dx)[XY^3 + XY] = (d/dx)[5]
Step 2: Apply the product rule and chain rule in order to differentiate the terms on the left side of the equation.
Using the product rule for the term XY^3, we have:
(d/dx)(XY^3) = Y^3(d/dx)(X) + X(d/dx)(Y^3)
Similarly, for the term XY, we have:
(d/dx)(XY) = Y(d/dx)(X) + X(d/dx)(Y)
Step 3: Differentiate the terms with respect to x using the chain rule.
Since Y is a function of x, we use the chain rule to differentiate it:
(d/dx)(Y) = dY/dx = Y' (Note: Y' means the derivative of Y with respect to x)
Step 4: Simplify the equation by substituting the values obtained from differentiating the terms.
Plugging in the values obtained from the chain rule, we have:
Y^3 * 1 + X * 3Y^2 * Y' + Y * 1 + X * Y' = 0
Rearranging the equation, we get:
X * 3Y^2 * Y' + X*Y' = -Y^3 - Y
Step 5: Solve for Y' to find the derivative of Y with respect to X.
Factoring out Y' from both terms on the left side of the equation, we have:
X * Y' * (3Y^2 + 1) = -Y^3 - Y
Dividing both sides by (3Y^2 + 1), we get:
Y' = (-Y^3 - Y) / (X * (3Y^2 + 1))
Step 6: Substitute the given point (3,2) into the equation obtained in step 5 to find the slope of the tangent line.
Plugging in X = 3 and Y = 2, we have:
Y' = (-2^3 - 2) / (3 * (2^2 + 1))
Simplifying, we get:
Y' = (-8 - 2) / (3 * (4 + 1))
= (-10) / (3 * 5)
= -10 / 15
Reducing the fraction, we get:
Y' = -2 / 3
Therefore, the slope of the tangent line to the curve XY^3 + XY = 5 at the point (3,2) is -2/3.