Find d^2x/dt^2 as a function of x if dx/dt=4xcosx.
take derivative with respect to t
=4(cos(x)dx/dt-xsin(x)dx/dt)
plug in 4xcosx for every dx/dt
=4(cos(x)(4xcosx)-xsin(x)(4xcosx))
Well, well, well, looks like we have some calculus fun in store! Let's break out those derivatives hats and get started, shall we?
Given that dx/dt = 4xcos(x), we need to find d^2x/dt^2. To do this, we'll invoke the magic of the chain rule, my favorite calculus party trick!
The chain rule states that if y = f(g(x)) then dy/dx = f'(g(x)) * g'(x) – the derivative of the outside function times the derivative of the inside function.
So, let's apply this rule like a tuxedo on a penguin to our equation: dx/dt = 4xcos(x). We have our outside function, cos(x), so we'll differentiate it with respect to x first, giving us -sin(x) (imagine the sad face of sine).
Next, we differentiate the inside function, 4x, with respect to t (remember, we're differentiating with respect to time, not x this time!). The derivative of 4x with respect to t is simply 4(dx/dt) (imagine a 4-leaf clover on a rollercoaster).
Finally, we multiply the two derivatives together: d^2x/dt^2 = 4(dx/dt) * (-sin(x)). Replacing dx/dt with our given value, we get:
d^2x/dt^2 = 4(4xcos(x)) * (-sin(x))
Now, isn't that a lovely function of x? Quite the mathematical juggling act, I must say! Keep practicing those derivatives, and remember, math is always more fun with a touch of humor!
To find the second derivative of x with respect to t (d^2x/dt^2) in terms of x, we need to differentiate the given expression dx/dt = 4xcos(x) with respect to t.
Let's start by using the product rule to differentiate the expression:
d/dt (dx/dt) = d/dt (4xcos(x))
Applying the product rule, we have:
d^2x/dt^2 = 4 * (d/dt (xcos(x))) + x * (d/dt (4cos(x)))
The derivative of xcos(x) can be found using the product rule again:
d/dt (xcos(x)) = x * (d/dt (cos(x))) + cos(x) * (d/dt (x))
The derivative of cos(x) with respect to t is 0 because cos(x) does not depend on t. The derivative of x with respect to t is dx/dt.
Therefore, we have:
d/dt (xcos(x)) = x * 0 + cos(x) * dx/dt = cos(x) * dx/dt
Substituting this back into our earlier expression, we get:
d^2x/dt^2 = 4 * (cos(x) * dx/dt) + x * (d/dt (4cos(x)))
Simplifying further:
d^2x/dt^2 = 4cos(x) * dx/dt + 4x * (-sin(x) * dx/dt)
Finally, substituting dx/dt = 4xcos(x) into the expression:
d^2x/dt^2 = 4cos(x) * (4xcos(x)) + 4x * (-sin(x) * (4xcos(x)))
Simplifying the expression further:
d^2x/dt^2 = 16x^2cos^2(x) - 16x^2cos(x)sin(x)
So, the second derivative of x with respect to t, as a function of x, is given by d^2x/dt^2 = 16x^2cos^2(x) - 16x^2cos(x)sin(x).
To find the second derivative of x with respect to t, you need to differentiate the given expression for dx/dt = 4xcos(x) with respect to t.
First, let's differentiate both sides of the equation with respect to t. Remember that x is a function of t, so we need to use the chain rule when differentiating x.
d/dt(dx/dt) = d/dt(4xcos(x))
To differentiate dx/dt with respect to t, we use the chain rule:
d/dt(dx/dt) = d/dx(4xcos(x)) * dx/dt
Differentiating 4xcos(x) with respect to x gives:
d/dx(4xcos(x)) = 4cos(x) - 4xsin(x)
Then multiply the result by dx/dt:
d/dt(dx/dt) = (4cos(x) - 4xsin(x)) * dx/dt
Substituting the given expression for dx/dt = 4xcos(x):
d/dt(dx/dt) = (4cos(x) - 4xsin(x)) * 4xcos(x)
Simplifying further:
d^2x/dt^2 = (16xcos^2(x) - 16x^2cos(x)sin(x))
Therefore, the second derivative of x with respect to t, as a function of x, is given by:
d^2x/dt^2 = 16xcos^2(x) - 16x^2cos(x)sin(x)
d²x/dt²
=d(dx/dt)/dt
=d(4xcosx)/dt
Use the product rule,
:::d(uv)=udv+vdu:::
=4(1*cos(x)+x(-sin(x)))
=4(cos(x)-xsin(x))