Part b is asking for the derivative of dy/dx with respect to t, which can be written as d/dt(dy/dx). In other words, it is asking for the rate of change of the derivative dy/dx with respect to t. To find this, we need to use the chain rule.
To solve both parts, we'll start by finding dy/dx, and then differentiate with respect to t using the chain rule.
a) To find dy/dx, we need to express y in terms of x and differentiate:
Given: x = t^2 + t and y = sin(t)
Rearranging the given equations:
t^2 + t = x
sin(t) = y
We need to eliminate t and express y in terms of x. We can do this by using the inverse of each equation:
t = (-1 ± √(1 + 4x))/2 (quadratic formula)
sin(t) = y
Since there are two possible values for t, we can write:
y = sin[(-1 ± √(1 + 4x))/2]
Now, differentiate y with respect to x using the chain rule:
dy/dx = (dy/dt)/(dx/dt)
Remember, we can express dy/dt as cos(t) since the derivative of sin(t) with respect to t is cos(t).
dy/dx = cos[(-1 ± √(1 + 4x))/2] / (d/dt[(−1 ± √(1 + 4x))/2] / d/dx[t^2 + t])
We can simplify d/dt[(−1 ± √(1 + 4x))/2] to:
d/dt[(−1 ± √(1 + 4x))/2] = (d/dt[-1 ± √(1 + 4x)]) / 2
Now, differentiate t^2 + t with respect to x, we obtain:
d/dx[t^2 + t] = (d/dt[t^2 + t]) / (d/dx[t^2 + t]) = 1 / (1 + 2t).
Substituting the above derivatives, we get:
dy/dx = cos[(-1 ± √(1 + 4x))/2] / [(d/dt[-1 ± √(1 + 4x)]) / 2] * (1 / (1 + 2t))
Note: We have not simplified the expression further because it depends on the sign chosen in the quadratic formula.
b) To find d/dt(dy/dx), we will further differentiate the expression obtained in part a:
d/dt(dy/dx) = d/dt[cos[(-1 ± √(1 + 4x))/2] / [(d/dt[-1 ± √(1 + 4x)]) / 2] * (1 / (1 + 2t))]
This involves differentiating both the numerator and denominator, and then applying the chain rule to obtain the final expression. The process can become quite complex, so it's best to use a symbolic differentiator or a computer algebra system to compute this derivative accurately.