Find dy/dx if y=cost, x=sint
AAAaannndd the bot gets it wrong yet again!
x^2+y^2 = 1
2x + 2yy' = 0
y' = -x/y
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.
To find dy/dx, we need to differentiate y with respect to x. Given that y = cos(t) and x = sin(t), we need to find dy/dx.
To do this, we need to use the chain rule. The chain rule states that if we have a composition of functions, such as y = f(g(x)), then the derivative of y with respect to x is given by dy/dx = f'(g(x)) * g'(x).
In our case, we have y = cos(t) and x = sin(t). To express y in terms of x, we can substitute sin(t) for x in y = cos(t). This gives us:
y = cos(t) = cos(arcsin(x)).
To find dy/dx, we need to differentiate cos(arcsin(x)) with respect to x. Let's proceed step by step:
1. Let's differentiate the outer function, cos(u), where u = arcsin(x). The derivative of cos(u) with respect to u is -sin(u):
d/dx [cos(u)] = -sin(u).
2. Next, let's find the derivative of the inner function, u = arcsin(x). The derivative of arcsin(x) with respect to x is 1/sqrt(1 - x^2):
d/dx [arcsin(x)] = 1 / sqrt(1 - x^2).
3. Now, let's substitute back the values of u and du/dx into the chain rule formula:
dy/dx = d/dx [cos(arcsin(x))] = -sin(arcsin(x)) * (1 / sqrt(1 - x^2)).
4. Simplifying further, we can recognize that sin(arcsin(x)) is equivalent to x. Therefore, substituting back, we have:
dy/dx = -x / sqrt(1 - x^2).
So, the derivative dy/dx is given by -x / sqrt(1 - x^2).
To find dy/dx, we can use the chain rule.
First, let's differentiate y = cos(t) with respect to t:
dy/dt = -sin(t).
Next, let's differentiate x = sin(t) with respect to t:
dx/dt = cos(t).
Now, we can find dy/dx by dividing dy/dt by dx/dt:
dy/dx = (dy/dt) / (dx/dt).
Substituting the derivatives we found earlier:
dy/dx = (-sin(t)) / (cos(t)).
However, we need to express our answer in terms of x.
Using the identity cos^2(t) + sin^2(t) = 1, we can express sin(t) in terms of x:
sin(t) = x.
Now, let's find cos(t):
cos(t) = sqrt(1 - sin^2(t)).
Since y = cos(t), we have:
y = sqrt(1 - x^2).
Finally, we can substitute cos(t) and sin(t) in terms of x into our expression for dy/dx:
dy/dx = (-sin(t)) / (cos(t)) = -x / sqrt(1 - x^2).
Therefore, the derivative dy/dx is given by -x / sqrt(1 - x^2).
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.