Find dy/dx if y=cos2t, x=sint
To find dy/dx, we can use the chain rule. Let's start by expressing y in terms of x:
Since x = sin(t), we can solve for t in terms of x: t = arcsin(x).
Next, we can express y = cos(2t) in terms of t by substituting the value of t: y = cos(2arcsin(x)).
Now we differentiate y with respect to x using the chain rule:
dy/dx = (dy/dt)(dt/dx)
Differentiating y with respect to t, we get:
dy/dt = -2sin(2arcsin(x))
Differentiating t with respect to x, we get:
dt/dx = dt/dt * dt/dx = 1 * cos(t) = cos(arcsin(x))
Substituting these values back into the equation dy/dx = (dy/dt)(dt/dx), we have:
dy/dx = -2sin(2arcsin(x)) * cos(arcsin(x))
Simplifying further, we can rewrite sin(2arcsin(x)) in terms of x as 2x * √(1 - x^2):
dy/dx = -2 * 2x * √(1 - x^2) * cos(arcsin(x))
Finally, we can simplify this expression:
dy/dx = -4x * √(1 - x^2) * cos(arcsin(x))
Therefore, the derivative of y = cos(2t) with respect to x when x = sin(t) is given by dy/dx = -4x * √(1 - x^2) * cos(arcsin(x)).
And we have another botched attempt by the bot
y=cos2t, x=sint
dy/dt = -2 sin 2t , dx/dt = cos t
dy/dx = (dy/dt) / dx/dt = -2sin 2t/cos t
= -4 sint cost/cost
= -4sint
Apologies for the mistake in the previous response. You are correct. Let's correct the calculation:
Given:
y = cos(2t)
x = sin(t)
To find dy/dx, we can differentiate both sides of the equations with respect to t:
dy/dt = -2sin(2t)
dx/dt = cos(t)
Now, we can express dy/dx in terms of dy/dt and dx/dt:
dy/dx = (dy/dt) / (dx/dt)
= (-2sin(2t)) / cos(t)
= -2sin(2t) / cos(t)
Since we have x = sin(t), we can substitute 2t = arcsin(x):
dy/dx = -2sin(arcsin(x)) / cos(t)
= -2x / cos(t)
= -2x / √(1 - x^2) (using the trigonometric identity: cos(t) = √(1 - sin^2(t)) = √(1 - x^2))
Therefore, dy/dx = -2x / √(1 - x^2).
What a lot of work!
dy/dx = (dy/dt) / (dx/dt) = -2sin2t/cost = -4sint = -4x
note that this is the same answer as Bot came up with, since
cos(arcsin(x)) = √(1-x^2)
also, since y = cos2t = 1 - 2sin^2t = 1 - 2x^2
that means that dy/dx = -4x
Apologies for the confusion in the previous response. You are correct. The correct answer is indeed dy/dx = -4x. Thank you for catching that error.
To find dy/dx, we need to find the derivative of y with respect to x. Since y is given as a function of t and x is also given as a function of t, we can use the chain rule to find dy/dx.
Using the chain rule, we have:
dy/dx = (dy/dt) / (dx/dt)
To find dy/dt, we take the derivative of y = cos(2t) with respect to t:
dy/dt = d/dt (cos(2t))
Since the derivative of cosine is negative sine, we get:
dy/dt = -sin(2t)
To find dx/dt, we take the derivative of x = sin(t) with respect to t:
dx/dt = d/dt (sin(t))
Since the derivative of sine is cosine, we get:
dx/dt = cos(t)
Now we can substitute these derivative expressions into our original equation for dy/dx:
dy/dx = (dy/dt) / (dx/dt)
= (-sin(2t)) / (cos(t))
So, dy/dx = -sin(2t) / cos(t).
To find dy/dx, we first need to express y in terms of x.
Given that x = sin(t), we can use the Pythagorean identity sin^2(t) + cos^2(t) = 1, to solve for cos(t) in terms of x.
Since sin^2(t) = 1 - cos^2(t), we can rewrite x^2 = sin^2(t) as x^2 = 1 - cos^2(t).
Rearranging the equation, we have cos^2(t) = 1 - x^2.
Taking the square root on both sides, we get cos(t) = ±√(1 - x^2).
Since y = cos(2t), we can substitute cos(t) in terms of x into the equation for y:
y = cos(2t) = cos^2(t) - sin^2(t) = (1 - x^2) - (1 - cos^2(t)) = 1 - x^2 - (1 - (1 - x^2)) = 1 - x^2 - (1 - 1 + x^2) = 1 - x^2 - x^2 = 1 - 2x^2.
Now we can find dy/dx by differentiating y = 1 - 2x^2 with respect to x:
dy/dx = d/dx (1 - 2x^2) = 0 - 4x = -4x.
Therefore, dy/dx = -4x.