What is the derivative of Sin(ArcCosx)?

Question

please help.

Answer 1

f(x) = sin(arccos(x))

One immediate instinct is to differentiate using the chain rule. I personally think it's a bad idea, because we can evaluate this prior to differentiating.

Let t = arccos(x). Then
cos(t) = x, so
cos(t) = x/1

Use the rules for SOHCAHTOA and think of right angle triangles.
cos(t) = adj/hyp = x/1, so

adj = x
hyp = 1, so by Pythagoras
opp = sqrt(1 - x^2)

Therefore

sin(t) = sin (arccos(x)) = opp/hyp = sqrt(1 - x^2)/1

f(x) = sqrt(1 - x^2). Differentiating, using the fact that
d/dx sqrt(x) = 1/[2sqrt(x)], we get

f'(x) = 1/[2sqrt(1 - x^2)] {-2x}
f'(x) = -2x/[2sqrt(1 - x^2)]
f'(x) = -x/sqrt(1 - x^2)

Answer 2

First let y=arccos x. Then cos y=x, so sin y = ±√(1-x²). However, the range of the (principal value of the) arccos function is [0, π], over which sin y is always nonnegative, so in fact sin (arccos x) = √(1-x²). Now differentiate:

d(√(1-x²))/dx = 1/(2√(1-x²)) * d(1-x²)/dx = 1/(2√(1-x²)) * (-2x) = -x/√(1-x²).

Alternatively, if you have memorized that the derivative of arccos x is -1/√(1-x²), you could proceed directly as follows:

d(sin (arccos x))/dx = cos (arccos x) * d(arccos x)/dx = x*d(arccos x)/dx = x*(-1/√(1-x²)) = -x/√(1-x²), which is what we obtained above.

Answer 3

chain rule:

take the derivative of the outside leaving the inside alone:

cos(ArcCosx)

then multiply that by the derivative of the inside

cos(ArcCosx) (-1/sqrt(1-x^2))

then cos(ArcCosX) cancel out to equal x so you're left with:

(-x/sqrt(1-x^2))

-- negative x over the square root of one minus x squared

Answer 4

Use the chain rule.

d[sin(arccos x)] / dx

= cos(arccos x)*d[arccos x]/dx

= x*d[arccos x]/dx

= x*[-1/√(1- x²)] = -x / √(1- x²)