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taurus [48]
3 years ago
8

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

Mathematics
1 answer:
sergey [27]3 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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It is given that the total measurement of the two angles combined would equate to 116°.

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m∠1 + m∠2 = 116°

m∠1 = m∠2 + 10°

First, plug in "m∠2 + 10" for m∠1 in the first equation:

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~

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