Find the gradient of the curve xy^3 = 5 ln y at the point (0,1).
take the differential:
3xy^2 dy+ y^3 dx=5/y dy
dy(2xy^2-5/y)=-y^3 dx
dy/dx= you finish it.
To find the gradient of the curve at a given point, we need to find the derivative of the curve with respect to x and substitute the coordinates of the point into the derivative.
Step 1: Differentiate the equation with respect to x using the product rule.
Let's differentiate the equation xy^3 = 5ln y.
Taking the derivative of both sides with respect to x, we get:
(y^3)(dx/dx) + (3xy^2)(dy/dx) = 5(dy/dx)(1/y)
Step 2: Simplify the equation by substituting the given coordinates (0, 1).
At the point (0, 1), x = 0 and y = 1.
Substituting these values into the equation, we get:
(1^3)(dx/dx) + (3(0)(1^2))(dy/dx) = 5(dy/dx)(1/1)
Simplifying further, we get:
(1)(1) + (0)(0) = 5(dy/dx)
1 = 5(dy/dx)
Divide both sides by 5:
dy/dx = 1/5
So, the gradient of the curve xy^3 = 5 ln y at the point (0, 1) is 1/5.