Find (dy)/(dx) and (d^2y)/(dx^2), and find the slope and...

Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$, and find the slope and concavity (if possible) at the given value of the parameter. (If an answer does not exist, enter DNE.) Parametric Equations $x = \sqrt{t}$, $y = 4t - 1$ Point $t = 4$ $\frac{dy}{dx} = $ $\frac{d^2y}{dx^2} = $ slope concavity: \text{---Select---} concave upward concave downward neither

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00:01 So in this question, they say find dydx and d squared y over dx squared, and then find the slope and concavity, if possible, at the given value of the parameter.

00:13 So first, i want to find dydx, not a, say, numerical value, but in terms of t.

00:22 And so let's see, how do we get dydx? well, dydx is equal to the derivative of y, with respect to t, divided by the derivative of x with respect to t.

00:41 Now, what is my derivative of y with respect to t? that would just be 2t.

00:51 While my derivative of x with respect to t is negative 1 over t squared, that's the derivative of 1 over t.

01:01 Now, if i am dividing by negative 1 over t squared, that's the same as multiplying by the reciprocal.

01:10 I'm multiplying by negative t squared to get negative 2 t cubed.

01:16 So my answer here is negative 2 t cubed.

01:22 Now, your slope was correct because if i evaluate d, y, dx at t equals negative 3, i get negative 2.

01:35 Times negative 3 being cubed.

01:39 Negative 2 times negative 27 is positive 54.

01:46 Now let's get a second derivative.

01:48 Okay.

01:51 D squared y over dx squared.

01:56 So what's my formula for this? i take the derivative with respect to t of dy, dx, and then i redivide that by dx d, t.

02:12 So my dydx was negative 2t cubed...

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