Prove that y = arcsinx on the root of 1 x squared satisfies the equality (1 - x squared)y' - xy = 1

Answer:

Step-by-step explanation:

The first part is a computation, combining the chain and quotient rules. We know that the derivative of f(x)=arcsin(x) is 1/sqrt(1-x2), and we need to differentiate the denominator. We use the chain rule: let u=1-x^2, and v=sqrt(u). Then du/dx=-2x, and dv/du = 1/(2 sqrt(u)). Combining, dv/dx=-x/sqrt(1-x2). Now, we put everything together using the quotient rules: y' = (f'v-v'f)/v2. Multiplying by (1-x2)=v2, we obtain (1-x2)y'= sqrt(1-x2)/sqrt(1-x2) +xarcsin(x)/sqrt(1-x2), which we can rewrite as 1+xy, as desired. Let's call this equation ().

We prove the second part by induction on n. To do this, we need two things: a base case, that is, proving the case n=0, and an inductive step, where we prove case n+1 from case n. To do the base case, we differentiate both sides of equation (*) that we proved above: d/dx ((1-x2 )y') = d/dx (xy+1 ). On the left-hand side, we get -2xy'+(1-x2 )y'', using the product rule, and on the right-hand side, we get xy'+y, similarly. Rearranging, we've proven the case n=0 of the desired statement. The inductive step is similar: we differentiate the equation from case n using the chain rule. Each of the first two terms splits into two: respectively, -2xy(n+2)+(1-x2)y(n+3), and (2n+3)y(n+1)+(2n+3)xy(n+2). If we combine the terms with the same numbers of derivatives on y, we get (1-x2)y(n+3)-(2n+5)xy(n+2)-((n+1)2+2n+3)y(n+1)=0 -which is what we wanted to prove, since ((n+1)2+2n+3) simplifies to (n+2)2.

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(−3 6/7)⋅(−3 1/3)

What is the product?

Enter your answer as a simplified mixed number in the box.

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167.67 divided by 8.1

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13. For a weekly car rental, a car-rental company charges customers a flat fee of $100 plus
10 cents for every mile traveled.
15 Write an equation for the amount, A dollars, of a weekly rental for a car that travels
m miles.
16
Create a table of m and A values using the equation from part a for 150, 200, and
250 miles.

Series
What number should replace
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Question 2 options:
True
False

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In your computations round all amounts to two decimal places. Round your final answer to whole number

4) 4+ 1 =
a) 5
b) 572
12
c) 51
d) 572
6
12

I need help solve the equation

he population of a particular town is given by the function P(t) = 1,200(1.05)2t, where t is the time in years and P(t) is the population after t years. What is the current population, the percentage growth rate, and the population size (rounded to the nearest whole person) in 7 years?

Hint: Percentage Growth Rate = r ⋅ 100 in A = A0(1 + r)t (4 points)

1,200, 5% annually, and 1,689
1,200, 50% semiannually, and 20,503
1,200, 10.5% per year, and 2,414
1,200, 5% semiannually, and 2,376

SOLVE PLZ I give brainlest
17+ -9

A. 26

B. 8

C. -8

D.-26

whats the answer can someone answer please​

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Show all your working.
2
3
Income tax
Everyone can earn a certain amount of money without paying tax. This
is called a Personal Allowance. They must pay tax on any earnings
over this allowance.
4
5
6
£11 850
Income tax Personal Allowance, 2018/2019
7
8
This formula gives the amount of Income tax a person pays in a year
T = 0.2 (y-p)
9
10
where T = income tax for the year
y = money earned per year
p = Personal Allowance
11
12
13
14
15
A caterer earns £1375 per month.

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Bella conducted a scientific experiment. For a certain time, the temperature of a compound rose
3
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degrees in
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15 minutes. What was the rate, in degrees per hour, that the temperature of the compound rose?

If p=7, then what is the value of 8×p−68×p-6?
A
6

B
9

C
50

D
56

Solve for g.
16g–11g+g=6
g=

Solve for u.
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7u–u+2u+11u=10
u=

A 95% confidence interval for the true proportion of teenagers who have been to a professional sporting event is
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O Yes, because 0.3525 is in the interval.
O Yes, because the entire interval is greater than 0.25.
O No, because there are values in the interval greater than 0.3525.
O No, because the center of the interval is 0.3525, which is not statistically greater than 0.25.

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