The derivative of the function is y' = (27x² ln(x) - 9x²) / (ln(x))²
Given data:
To find the derivative of the function y = (9x³) / ln(x), we can use the quotient rule.
The quotient rule states that if we have a function in the form f(x) / g(x), where f(x) and g(x) are differentiable functions, the derivative is given by:
(f'(x) * g(x) - f(x) * g'(x)) / (g(x))²
Let's apply the quotient rule to the given function:
f(x) = 9x³
g(x) = ln(x)
f'(x) = 27x² (derivative of 9x³ with respect to x)
g'(x) = 1/x (derivative of ln(x) with respect to x)
Now we can substitute these values into the quotient rule formula:
y' = ((27x²) * ln(x) - (9x³) * (1/x)) / (ln(x))²
Simplifying further:
y' = (27x² ln(x) - 9x²) / (ln(x))²
Hence , the derivative of the function y = (9x³) / ln(x) is:
y' = (27x² ln(x) - 9x²) / (ln(x))²
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Given the sequence defined as follows: an=√an−1+2,n≥1,a0=1. Which properties does this sequence possess? a) The sequence is increasing and unbounded. b) The sequence is increasing and bounded above by 2 . c) The sequence is decreasing and bounded below by 1 . d) The sequence diverges.
The answer is (b) The sequence is increasing and bounded above by 2.
To determine the properties of the given sequence, let's examine its behavior. Starting with a₀ = 1, we can generate the terms of the sequence:
a₁ = √(a₀) + 2 = √(1) + 2 = 3
a₂ = √(a₁) + 2 = √(3) + 2 ≈ 3.732
a₃ = √(a₂) + 2 ≈ 3.732
...
From the pattern observed, we can conclude that the sequence is increasing. Each term is larger than the previous one, as the square root and addition of 2 will always result in a larger value.
To determine if the sequence is bounded, we can examine its behavior as n approaches infinity. As n increases, the terms of the sequence approach a limit. Let's assume this limit is L. Taking the limit of both sides of the recursive formula, we have:
L = √(L) + 2
Solving this equation, we get L = 2. Thus, the sequence is bounded above by 2.
In summary, the sequence is increasing, as each term is larger than the previous one. Additionally, the sequence is bounded above by 2, as it approaches the limit of 2 as n approaches infinity. Therefore, the correct answer is (b) The sequence is increasing and bounded above by 2.
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Suppose u(0)=0 and u′(0)=98. If (u/q)′(0)=7, what is q(0) ?
We found that q(0) = 14. By applying the quotient rule and evaluating the expression at x = 0, we obtained an equation that allows us to solve for q. Dividing both sides by q and simplifying, we found that q = 14.
Let's start by using the quotient rule to find the derivative of u/q. The quotient rule states that for two functions u(x) and q(x), the derivative of their quotient is given by:
(u/q)' = (u'q - uq') / q^2
We are given that (u/q)'(0) = 7. Substituting this value into the quotient rule, we have:
(u'q - uq') / q^2 = 7
At x = 0, we can evaluate the expression further. We are also given that u(0) = 0 and u'(0) = 98. Substituting these values into the equation, we have:
(98q - 0) / q^2 = 7
Simplifying the equation, we have:
98q = 7q^2
Dividing both sides by q, we have:
98 = 7q
Solving for q, we find:
q = 98 / 7 = 14
Therefore, q(0) = 14.
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Verify Green's Theorem in the plane for F = xy i + x^2 j and where C is the boundary of the region between the graphs of y = x^2 and y = 1.
Verify Stokes's Theorem for F = (z – y) i + (x − z)j+(x − y)k and S : z = √1−x^2−y^2. Assume outward normal.
Verify Gauss Divergence Theorem for F = xy^2 i + yx^2 j + ek^ for the solid region D bounded by z = √x^2+y^2 and z = 4.
Green's Theorem can be defined as the relationship between a line integral and a double integral over a plane region. It is named after the mathematician George Green. Gauss's Divergence Theorem can be defined as the relationship between a flux integral and a triple integral over a region.
Here, we need to verify Green's Theorem in the plane for the vector field F = xy i + x² j and where C is the boundary of the region between the graphs of y = x²
and y = 1.
Stokes's Theorem states that a line integral of a vector field around a closed loop is equal to a surface integral of the curl of the vector field over the surface bounded by the loop. The surface integral of the curl over S is∫∫S (curl F).dS= ∫∫S (-2i - 2j - 2k).(√(2 - 2x² - 2y²) dA)= -2 ∫∫S √(2 - 2x² - 2y²) dA We can change to polar coordinates to evaluate the integral. In polar coordinates, the integral becomes∫(0,2π)∫(0,1) √(2 - 2r²) r drdθ= 2π/3
Hence, by Stokes's Theorem, the line integral of F around any closed curve C in S is equal to -2π/3 times the area enclosed by C. Gauss's Divergence Theorem can be defined as the relationship between a flux integral and a triple integral over a region. Here, we need to verify the Gauss Divergence Theorem for F = xy² i + yx² j + ek for the solid region D bounded by z = √(x² + y²) and z = 4.The divergence of F is∇. F = (∂P/∂x + ∂Q/∂y + ∂R/∂z)
= y² + x²
Since the region D is a solid, we need to use the divergence theorem in its integral form:∫∫S F.N dS = ∫∫∫D ∇.F dV Here, S is the surface of the solid D and N is the outward unit normal vector to S. the Gauss Divergence Theorem is verified for the given F and region D.
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5. (2 points) Evaluate the following integrals. a. \( \int\left(3 x^{4}-7 x+2\right) d x \) b. \( \int\left(\frac{24 x^{4}-9-6 x}{3 x}\right) d x \)
A[tex])\[\int\left(3 x^{4}-7 x+2\right) d x=\frac{3}{5}x^5-\frac{7}{2}x+2x+C\][/tex]where C is a constant of integration.
B[tex])\[\int\left(\frac{24 x^{4}-9-6 x}{3 x}\right) d x=2x^4-5\ln|x|+C\][/tex]where C is a constant of integration.
a. [tex]\(\int\left(3 x^{4}-7 x+2\right) d x\)[/tex]
Here, we use the sum rule of integration.
The integral of a sum is the sum of the integrals. So,\[tex][\int(3x^4-7x+2)dx=\int3x^4dx-\int7xdx+\int2dx\]\[=\frac{3}{5}x^5-\frac{7}{2}x+2x+C\][/tex]
Therefore,
[tex]\[\int\left(3 x^{4}-7 x+2\right) d x=\frac{3}{5}x^5-\frac{7}{2}x+2x+C\][/tex]
where C is a constant of integration.
b. [tex]\(\int\left(\frac{24 x^{4}-9-6 x}{3 x}\right) d x\)[/tex]
First, simplify the fraction:
[tex]\[\frac{24x^4-9-6x}{3x}=8x^3-3-\frac{2}{x}\][/tex]
Now, integrate each term separately. Recall that the integral of 1/x is[tex]ln|x|.[/tex]
Thus,[tex]\[\int\left(\frac{24 x^{4}-9-6 x}{3 x}\right) d x=\int8x^3 dx - \int3 dx-\int\frac{2}{x}dx\]\[=2x^4-3\ln|x|-2\ln|x|+C\][/tex]
Therefore,
[tex]\[\int\left(\frac{24 x^{4}-9-6 x}{3 x}\right) d x=2x^4-5\ln|x|+C\][/tex]
where C is a constant of integration.
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If the 4th and 7th terms of a geometric sequence are 1/16 and
1/128, then the sum of the first 7 terms of this sequence is equal
to
Therefore, the sum of the first 7 terms of the given geometric sequence is 127/128.
To find the sum of the first 7 terms of a geometric sequence, we need to determine the common ratio and the first term of the sequence.
Let's denote the first term of the sequence as 'a' and the common ratio as 'r'.
Given that the 4th term is 1/16 and the 7th term is 1/128, we can write the following equations:
a * r^3 = 1/16 (equation 1)
a * r^6 = 1/128 (equation 2)
Dividing equation 2 by equation 1, we get:
(r^6)/(r^3) = (1/128)/(1/16)
r^3 = 1/8
Taking the cube root of both sides, we find:
r = 1/2
Substituting the value of r back into equation 1, we can solve for 'a':
a * (1/2)^3 = 1/16
a * 1/8 = 1/16
a = 1/2
Now we have the first term 'a' as 1/2 and the common ratio 'r' as 1/2.
The sum of the first 7 terms of the geometric sequence can be calculated using the formula:
Sum = a * (1 - r^n) / (1 - r)
Substituting the values into the formula, we have:
Sum = (1/2) * (1 - (1/2)^7) / (1 - 1/2)
Simplifying the expression
Sum = (1/2) * (1 - 1/128) / (1/2)
Sum = (1/2) * (127/128) / (1/2)
Sum = (127/128)
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which value of x results in short circuit evaluation, causing y < 4 to not be evaluated? (x >= 7) & (y < 4) a. 6 b. 7 c. 8 d. no such value
The value of x that results in short circuit evaluation, causing y < 4 to not be evaluated, is option c. 8.
In short circuit evaluation, the logical operators (such as "&&" in this case) do not evaluate the right-hand side of the expression if the left-hand side is sufficient to determine the final outcome.
In the given expression, (x >= 7) is the left-hand side and (y < 4) is the right-hand side. For short circuit evaluation to occur, the left-hand side must be false, as a false condition would make the entire expression false regardless of the right-hand side.
If we substitute x = 8 into the expression, we have (8 >= 7) & (y < 4). The left-hand side, (8 >= 7), evaluates to true. However, for short circuit evaluation to happen, it should be false. Hence, the right-hand side, (y < 4), will not be evaluated, and the final result will be true without considering the value of y. Thus, option c, x = 8, satisfies the condition for short circuit evaluation.
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