Question

Given the equation below, find dy/dx.

14x²+4x¹⁴y+y³=19
dy/dx = ____
Now, find the equation of the tangent line to the curve at (1,1). Write your answer in mx+b format.
y= ____

Answer 1

To find dy/dx, we differentiate the given equation implicitly with respect to x. At the point (1,1), the derivative dy/dx (the slope of the tangent) is -12. Thus, the equation of the tangent line at (1,1) is y = -12x + 13.

Step-by-step explanation:

To find dy/dx, we'll differentiate both sides of the equation with respect to x, treating y as an implicit function of x. The original equation is:

14x² + 4x¹⁴y + y³ = 19

When we differentiate both sides with respect to x, we get:

28x + 56x¹³y + (4x¹⁴)(dy/dx) + 3y²(dy/dx) = 0

Now we can solve this equation for dy/dx by isolating its terms on one side:

(4x¹⁴ + 3y²)dy/dx = -28x - 56x¹³y

Then dividing both sides by (4x¹⁴ + 3y²), we find:

dy/dx = (-28x - 56x¹³y) / (4x¹⁴ + 3y²)

To find the equation of the tangent line at the point (1,1), we substitute x = 1 and y = 1 into the equation of dy/dx to calculate the slope m:

m = (-28(1) - 56(1)¹³(1)) / (4(1)¹⁴ + 3(1)²) = -84 / 7 = -12

The tangent line will have the form y = mx + b. Because the line passes through the point (1,1), we can find b:

1 = (-12)(1) + b

b = 13

Therefore, the equation of the tangent line is y = -12x + 13.