Would the first derivative of

R(t)=9t[(t-4)^3/2.718] be R'(t)=3.311258278(3(t-4)^2(t)+(t-4)^3) and the second derivative be
R''(t)=3.311258278(6(t-4)x+6(t-4)^2)?

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Just type "derivative" or "2nd derivative" and your function. Check to be sure it interprets your input as you intended. Parentheses are your friend.

For your function, you correctly extracted the 9/e constant, and the derivative of t(t-4)^3 is
3(t-4)^2 + (t-4)^3 = (t-1)(t-4)^2

2nd derivative is 12(t-2)(t-4)

Your answers appear correct

To find the derivatives of the function R(t) = 9t[(t - 4)^(3/2.718)], we can apply the rules of differentiation. Let's break it down step by step:

Step 1: Simplify the expression inside the brackets.
(t - 4)^(3/2.718) can be rewritten as e^(3ln(t - 4)), where e is Euler's number. This step allows us to work with a simpler form.

Step 2: Apply the product rule to differentiate 9t * e^(3ln(t - 4)).

The product rule states that if we have two functions, u(t) and v(t), then the derivative of their product u(t) * v(t) is given by:
(u * v)' = u' * v + u * v'

Applying the product rule:
R'(t) = (9t)' * e^(3ln(t - 4)) + 9t * (e^(3ln(t - 4)))'

Step 3: Differentiate each term separately.

For (9t)', we use the power rule of differentiation, which states that the derivative of t^n, where n is a constant, is given by:
(d/dt)(t^n) = n * t^(n-1)

Using the power rule, (9t)' = 9.

For (e^(3ln(t - 4)))', we need to use the chain rule, which states that if we have a composite function f(g(t)), the derivative is given by:
(d/dt)(f(g(t))) = f'(g(t)) * g'(t)

Let u = 3ln(t - 4). Then applying the chain rule, we find that (e^(3ln(t - 4)))' = (e^u)' = e^u * u'

To find u', we differentiate 3ln(t - 4) using the chain rule:
u' = (d/dt)(3ln(t - 4)) = 3 * (1/(t - 4)) * (d/dt)(t - 4) = 3/(t - 4)

So, (e^(3ln(t - 4)))' = e^u * u' = e^(3ln(t - 4)) * (3/(t - 4))

Step 4: Substitute the derivative terms back into R'(t).

R'(t) = 9 * e^(3ln(t - 4)) + 9t * (e^(3ln(t - 4)) * (3/(t - 4)))

Simplifying further,
R'(t) = 9 * e^(3ln(t - 4)) + 27t * e^(3ln(t - 4)) / (t - 4)

Now, to find the second derivative R''(t), we differentiate R'(t) with respect to t.

R''(t) = (R'(t))'

Differentiating R'(t) involves applying the product rule again.

R''(t) = (9 * e^(3ln(t - 4)))' + (27t * e^(3ln(t - 4)))' / (t - 4) + 27t * e^(3ln(t - 4)) * (1/(t - 4))'

Simplifying further,
R''(t) = 0 + 0 + (27t * e^(3ln(t - 4)) * (1/(t - 4))'
R''(t) = 27t * e^(3ln(t - 4)) * (1/(t - 4))'

Using the chain rule, we differentiate (1/(t - 4)) as:
(d/dt)(1/(t - 4)) = -1 / (t - 4)^2

Substituting this back into R''(t),
R''(t) = 27t * e^(3ln(t - 4)) * (-1 / (t - 4)^2)

Finally, simplifying the expression:
R''(t) = -27t * e^(3ln(t - 4)) / (t - 4)^2

So, the first derivative R'(t) = 9 * e^(3ln(t - 4)) + 27t * e^(3ln(t - 4)) / (t - 4), and the second derivative R''(t) = -27t * e^(3ln(t - 4)) / (t - 4)^2.