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Dafna11 [192]
2 years ago
15

Use the inverse function-inverse cofunction

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

Answer:

arccot(x)'=-\frac{1}{1+x^2}

Step-by-step explanation:

<u>Use implicit differentiation:</u>

y=arccot(x)

cot(y)=x

\frac{1}{tan(y)}=x

\frac{cos(y)}{sin(y)}=x

\frac{dy}{dx}(\frac{cos(y)}{sin(y)})=\frac{dy}{dx}(x)

\frac{sin(y)(-sin(y))-cos(y)cos(y)}{sin^2(y)}*\frac{dy}{dx}=1 (Quotient Rule: \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2} )

\frac{-sin^2(y)-cos^2(y)}{sin^2(y)}*\frac{dy}{dx}=1

\frac{-(sin^2(y)+cos^2(y))}{sin^2(y)}*\frac{dy}{dx}=1

\frac{-1}{sin^2(y)}*\frac{dy}{dx}=1

-csc^2(y)\frac{dy}{dx}=1

\frac{dy}{dx}=\frac{1}{-csc^2(y)}

\frac{dy}{dx}=-sin^2(y)

\frac{dy}{dx}=-sin^2(arccot(x))

See the attached picture to understand how to evaluate the mixed composition of trig functions using a right triangle.

Therefore, the derivative of arccot(x) is -\frac{1}{\sqrt{1+x^2}}.

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We'll use standard labeling of right triangle ABC, C=90 degrees, legs a, b, hypotenuse c.

11.

Right triangle, cliff peak A, boat B, angle opposite cliff is B=28.9 deg. adjacent leg a=65.7 m, cliff height is leg b.

tan B = b/a

b = a tan B = 65.7 tan 28.9° = 36.3 m

12.

Similar story, boat at B, opposite b=3.5 m, rope c=12 m

sin B = b/c

B = arcsin b/c = arcsin (3.5/12) = 17.0°

13.

c=124 m, A=58°

sin A = a/c

a = c sin A = 124 sin 58 = 105.2 m

14.

That's a hypotenuse c=4-1.2 = 2.8 m to a height b=1.8m so

cos A = b/c

A = arccos b/c = arccos (1.8/2.8) = 50.0°

15.

Not a right triangle, an isosceles triangle.  Half of it is a right triangle with hypotenuse one arm, c=9.8 cm and angle opposite half the base of B=62/2=31°.  We're after d=2b:

sin B = b/c

b = c sin B

d = 2b = 2 c sin B = 2(9.8) sin 31 = 10.1 cm

Almost equilateral


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3 years ago
Find the edge length of the cube Volume = 0.064 m^3<br><br>​
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The edge length of the cube is 0.4
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2 years ago
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Pls help me this is 7th grade math
Ymorist [56]

Answer:

118 degrees

Step-by-step explanation:

180-62 = 118

8 0
2 years ago
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