Find dy/dx if y=cost, x=sint

Answer 1

To find dy/dx, we need to calculate the derivative of y with respect to x.

Given y = cos(t) and x = sin(t), we can express cos(t) in terms of x:

cos(t) = cos(arcsin(x))

To find dy/dx, we will differentiate both sides of the equation with respect to x:

d/dx (cos(t)) = d/dx (cos(arcsin(x)))

Using the chain rule, the right side becomes:

-sin(arcsin(x)) * d/dx (arcsin(x))

We know that sin(arcsin(x)) = x, so we can simplify the equation to:

-d/dx (arcsin(x))

The derivative of arcsin(x) with respect to x is:

d/dx (arcsin(x)) = 1 / sqrt(1 - x^2)

Therefore, dy/dx = -1 / sqrt(1 - x^2)

So, dy/dx = -1 / sqrt(1 - sin^2(t))

Notice that if we substitute sin(t) into the equation, we get:

dy/dx = -1 / sqrt(1 - sin^2(t))

Thus, dy/dx = -1 / sqrt(1 - x^2) for all values of x.

Answer 2

AAAaannndd the bot gets it wrong yet again!

x^2+y^2 = 1
2x + 2yy' = 0
y' = -x/y

Answer 3

Apologies for the previous incorrect response. Let's find dy/dx correctly this time.

Given the equation x^2 + y^2 = 1, we can differentiate both sides of the equation with respect to x:

2x + 2y * dy/dx = 0

Now, let's solve for dy/dx:

2y * dy/dx = -2x
dy/dx = -2x / (2y)
dy/dx = -x / y

Therefore, dy/dx = -x / y.