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taurus [48]
2 years ago
8

Using chain rule, what is the derivative of sin(arcsin(x))

Mathematics
1 answer:
sergey [27]2 years ago
7 0
Chain rule
y=f(g(x))
y´=(d f(gx)/d g)(d g/d x)

or

y=y(v)   and v=v(x), then dy /dx=(dy/dv)(dv/dx)

in our case:
y=sin (v)
v=arcsin(x)

dy/dv=d sin (v)/dv=cos (v)=cos(arcsin(x)
dv/dx=d arcsin(x)/dx=1/√(1-x²)

dy/dx=[cos (arcsin(x))]/√(1-x²)

Answer: d sin(arcsin(x))/dx=[cos (arcsin(x))]/√(1-x²)
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Equation of a circle:

The equation of a circle with center (x_0,y_0) is given by:

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This means that x_0 = -14, y_0 = -3

So

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This means that when x = -5, y = 3. We use this to find the radius squared. So

(x + 14)^2 + (y + 3)^2 = r^2

(-5 + 14)^2 + (3 + 3)^2 = r^2

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So, the equation of the circle is:

(x + 14)^2 + (y + 3)^2 = r^2

(x + 14)^2 + (y + 3)^2 = 117

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Step-by-step explanation:

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