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3241004551 [841]
2 years ago
9

Y=arccos(1/x) Please help me do them all! I don’t know derivatives :(

Mathematics
1 answer:
Stells [14]2 years ago
4 0

Answer:

f(x) =  {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x  \rightarrow \: x = sec \: y\\  \frac{dx}{dx}  =  \frac{d(sec \: y)}{dx}  \\ 1 = \frac{d(sec \: y)}{dx} \times  \frac{dy}{dy}  \\ 1 = \frac{d(sec \: y)}{dy} \times  \frac{dy}{dx}  \\1 = tan \: y.sec \: y. \frac{dy}{dx}  \\ \frac{dy}{dx} =  \frac{1}{tan \: y.sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ \sqrt{( {sec}^{2}   \: y - 1}) .sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\   \therefore  \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx}  =  \frac{1}{ |5x |  \sqrt{25 {x }^{2}  - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} }

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I really need help with this equation as soon as possible :D
viktelen [127]
The subtraction sign in front of the second set of parenthesis changes the signs on the terms inside those parenthesis.  So this could be rewritten as  x-3x^4+2x^2+x^3-4x^6.  None of the exponents on the x are the same, so we will just express it in terms of descending powers of x.   -4x^6-3x^4+x^3+2x^2+x.   This is also the same as  4x^6+3x^4-x^3-2x^2-x  if you don't like to start with a negative.  Either one is correct.
7 0
3 years ago
What is the equation of this graph? <br><br>a) x=-7<br>b) y=-7<br>​
sergiy2304 [10]
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4 0
3 years ago
Given that the square root function is only defined for non-negative numbers, define the domain of f in terms of c. {x | ≥ 0}
diamong [38]

Answer:

c:(0;+∞)

Step-by-step explanation:

Given the situation, that the square root function only allows you values above than "0" (not equal neither), then you must consider that every value above 0 belongs to it's domain.

Then, to express the domain, going from your most negative number, to your most possitive number (in this case all positive number, thats why we use infinite)  you must use the parenthesis wich means, you are not considering the value (in this case 0), but the value right after it, to the next value that as we said before, is inifinite. Also remember, that when you express a domain, and use infinite (despite it's going to negative way, or possitive way, it also goes with parenthesis).

5 0
3 years ago
The answer to this problem?
LenaWriter [7]

The simplified expression of \sqrt{\frac{162x^9}{2x^{27}}} is \frac{9}{x^9}

<h3>How to simplify the expression?</h3>

The expression is given as:

\sqrt{\frac{162x^9}{2x^{27}}}

Divide 162 and 2 by 2

\sqrt{\frac{81x^9}{x^{27}}}

Take the square root of 81

9\sqrt{\frac{x^9}{x^{27}}}

Apply the quotient rule of indices

9\sqrt{\frac{1}{x^{27-9}}}

Evaluate the difference

9\sqrt{\frac{1}{x^{18}}}

Take the square root of x^18

\frac{9}{x^9}

Hence, the simplified expression of \sqrt{\frac{162x^9}{2x^{27}}} is \frac{9}{x^9}

Read more about expressions at:

brainly.com/question/723406

#SPJ1

5 0
1 year ago
What is the length of side AB with endpointsA(–5, 3) and B(4, 3)? units
QveST [7]

Answer:

9 units

Step-by-step explanation:

the y value doesnt change, so the distance from -5 to 4 is 9 units

5 0
2 years ago
Read 2 more answers
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