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4 years ago

Derivative of x^2 •root x

2 answers:

6 0

Y= x².√x

y =x².(x)¹/² → y = x²⁺¹/² → y = x⁵/²

Remember that the derivative of y = xⁿ is y' = n. xⁿ⁻¹

y = x⁵/² → y' = (5/2).x⁽⁵/²⁾⁻¹ → y' = (5/2)x³/² or y' = (5/2)√x³ or y' = (5/2)x√x

butalik [34]4 years ago

4 0

= x^2 * x^1/2
= x^(5/2)

derivative = (5/2) x ^(5/2 - 1) = (5/2) x^(3/2) or (5/2)(sqrt x)^3

You might be interested in

Sam has $15 more dollars than Sarah, Casey has 2 times as much as Sam. The total they have all together is $95, how much money d

Alja [10]

Answers:

Sarah has $12.50
Sam has $27.50
Casey has $55

======================================================

Explanation:

x = amount of money Sarah has

x+15 = amount of money Sam has, since he has $15 more than her

2(x+15) = amount of money Casey has, since he has twice as much as Sam.

Let's add up those three expressions and set that sum equal to 95

Solve for x.

Sarah + Sam + Casey = 95

x+(x+15) + 2(x+15) = 95

x + x + 15 + 2x + 30 = 95

4x+45 = 95

4x = 95-45

4x = 50

x = 50/4

x = 12.50

This says that Sarah has $12.50

That must mean Sam has x+15 = 12.50+15 = 27.50 dollars

And Casey must have 2*27.50 = 55 dollars.

----------------

As a check,

Sarah+Sam+Casey = 12.50+27.50+55 = 95

So the answer is confirmed.

6 0

3 years ago

Read 2 more answers

In Chelsea's group, there are three boys and six girls. Write a fraction, in simplest form, to represent the amount of boys in t

DochEvi [55]

Answer:

1/2?

Step-by-step explanation:

3 boys over 6 girls. that would be 3/6. 3/6 is already 1/2. but its also simplest form.

6 0

4 years ago

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Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago

#13: The difference between two numbers is 3. Their sum if 27. What is the smaller number? * Your answer

Lelu [443]

Answer:

The smaller number is 12 & greater number is 15.

8 0

3 years ago

Read 2 more answers

+2+1+6 this is confusing what is the answer and explanation

SSSSS [86.1K]

Answer:

Step-by-step explanation:

hi I'm pretty sure this is really wrong but just kinda guessing

+2 is just a positive 2 so it's just known as 2

2+1+6 = 9

4 0

3 years ago

Read 2 more answers