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To compute cos(sin-1(x)) use the fact that. (cos(sin-1(x))). 2. + (sin(sin-1(x))) ... Now use that y = tan-1(x) to get a derivative entirely in terms of x.
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(1) Compute d dx
[sin−^1 (x)]. Let y = sin−^1 (x). Then
sin(y) =
Implicitly differentiate and solve for y′.
y′^ =
Now use that y = sin−^1 (x) to get a derivative entirely in terms of x
d dx
[sin−^1 (x)] = y′^ =
To compute cos(sin−^1 (x)) use the fact that ( cos(sin−^1 (x))
sin(sin−^1 (x))
Solve for cos(sin−^1 (x)). Remember you ±’s. The range of sin−^1 is. (Look it up if you need to.) On that set of inputs, cosine is (positive / negative)circle one
So that cos(sin−^1 (x)) =
And
d dx [sin
− (^1) (x)] =
1
(2) Compute d dx
[cos−^1 (x)]. Let y = cos−^1 (x). Then
cos(y) =
Implicitly differentiate and solve for y′.
y′^ =
Now use that y = cos−^1 (x) to get a derivative entirely in terms of x
d dx
[cos−^1 (x)] = y′^ =
To compute sin(cos−^1 (x)) use the fact that ( cos(cos−^1 (x))
sin(cos−^1 (x))
Solve for sin(cos−^1 (x)). Remember you ±’s. The range of cos−^1 is. (Look it up if you need to.) On that set of inputs, cosine is (positive / negative)circle one
So that sin(cos−^1 (x)) =
And d dx
[cos−^1 (x)] =
(4) Compute d dx
[sec−^1 (x)]. Let y = sec−^1 (x). Then
sec(y) =
Implicitly differentiate and solve for y′.
y′^ =
Now use that y = sec−^1 (x) to get a derivative entirely in terms of x
d dx
[sec−^1 (x)] = y′^ =
Above you see a trig function evaluated at an inverse trig function. Do the trigonometry needed to simplify. Your final answer should involve no trig functions.
So that d dx
[sec−^1 (x)] = y′^ =
And
d dx [tan
− (^1) (x)] =
