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Igoryamba
3 years ago
6

How to solve for first derivative?? I have two questions, one where I have the answer, however one where I don’t. The format of

the answer must be similar to the first question. But how to find first derivative for second question?

Mathematics
1 answer:
rjkz [21]3 years ago
5 0

Answer:

Q1: f'(x) = (-6x^8 + 2x^4 + 4)(48x^7) + (6x^8)(-48x^7 + 8x^3)

Q2: f'(x) = (6x^4)(18x^2 - 54x^8) + (6x^3 - 6x^9 + 3)(24x^3)

Step-by-step explanation:

The derivative of the product of two functions is:

f(x) = v(x)u(x)

f'(x) = v(x)u'(x) + u(x)v'x)

The derivative is the product of the first function and the derivative of the second function added to the product of the second function and the derivative of the first function.

Q1: The function you are given is:

f(x) = 6x^8(-6x^8 + 2x^4 + 4)

You can think of that function as the product of functions

u(x) = 6x^8 and v(x) = -6x^8 + 2x^4 + 4

We first find the derivatives of functions u and v:

u'(x) = 48x^7 and v'(x) = -48x^7 + 8x^3

Now we follow the rule above:

f'(x) = v(x)u'(x) + u(x)v'x)

f'(x) = (6x^8)(-48x^7 + 8x^3) + (-6x^8 + 2x^4 + 4)(48x^7)

Use the commutative property to change the order of the sum.

f'(x) = (-6x^8 + 2x^4 + 4)(48x^7) + (6x^8)(-48x^7 + 8x^3)

This is the solution you have.

Q2: The function you are given is:

f(x) = 6x^4(6x^3 - 6x^9 + 3)

You can think of that function as the product of functions

u(x) = 6x^4 and v(x) = 6x^3 - 6x^9 + 3

We first find the derivatives of functions u and v:

u'(x) = 24x^3 and v'(x) = 18x^2 - 54x^8

Now we follow the rule above:

f'(x) = v(x)u'(x) + u(x)v'x)

f'(x) = (6x^4)(18x^2 - 54x^8) + (6x^3 - 6x^9 + 3)(24x^3)

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KIM [24]

Answer:

  (5, 8), (7, 1), (15, 3)

Step-by-step explanation:

a) The variables are defined in the problem statement. The total expenditure must satisfy ...

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b) Possible solutions include ...

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3 years ago
Tyler went to the supermarket to buy food for a food pantry. He has $36, and can carry up to 20 pounds of food in his backpack.
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Answer:

Solutions: (2,10), (4,5)

Not solutions: (1,12), (6,10), (12,8), (18,6)

Step-by-step explanation:

Let x be the number of packages of pasta and y be the number of jars of pasta sauce. If pasta costs $1 for a 1-pound package, then x packages of pasta cost $x and weigh x pounds. If pasta sauce costs $3 for a 1.5 pound jar, then y jars cost $3y and weigh 1.5y pounds.

1. Tyler has $36, then

x+3y\le 36.

2. Tyler can carry up to 20 pounds of food in his backpack, then

x+1.5y\le 20.

You get the following system of inequalities:

\left\{\begin{array}{l}x+3y\le 36\\ x+1.5y\le 20\end{array}\right.

Now substitute the coordinates of each point:

<u>(1,12):</u>

\left\{\begin{array}{l}1+3\cdot 12=37> 36\\ 1+1.5\cdot 12=19\le 20\end{array}\right.

False, because first inequality doesn't hold.

<u>(2,10):</u>

\left\{\begin{array}{l}2+3\cdot 10=32\le 36\\ 2+1.5\cdot 10=17\le 20\end{array}\right.

True, both inequalities hold.

<u>(4,5):</u>

\left\{\begin{array}{l}4+3\cdot 5=19\le 36\\ 4+1.5\cdot 5=11.5\le 20\end{array}\right.

True, both inequalities hold.

<u>(6,10):</u>

\left\{\begin{array}{l}6+3\cdot 10=36\le 36\\ 6+1.5\cdot 10=21> 20\end{array}\right.

False, because secondt inequality doesn't hold.

<u>(12,8):</u>

\left\{\begin{array}{l}12+3\cdot 8=36\le 36\\ 12+1.5\cdot 8=24> 20\end{array}\right.

False, because second inequality doesn't hold.

<u>(18,6):</u>

\left\{\begin{array}{l}18+3\cdot 6=36\le 36\\ 18+1.5\cdot 6=27> 20\end{array}\right.

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7 0
3 years ago
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7 0
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Answer:

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<em />

Answer:  45.95x + 30.50

6 0
3 years ago
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