13 DERIVATIVES OF
|
y | = | arcsin x | |
implies | |||
1) sin y | = | x. | |
Therefore, according to the Pythagorean identity a': | |||
cos y | = | ||
= | |||
according to line 1). |
We take the positive sign because cos y is positive for all values of y in the range of y = arcsin x, which is the 1st and 4th quadrants. (Topic 19 of Trigonometry.)
Problem 2. If y = arcsec x, show:
Begin:
y | = | arcsec x | |
implies | |||
sec y | = | x. | |
Therefore, according to the Pythagorean identity b: | |||
tan y | = | ± | |
= | ± |
The derivative of y = arcsin x
The derivative of the arcsine with respect to its argument
is equal to 1 over the square root
of 1 minus the square of the argument.
Here is the proof:
y | = | arcsin x | |
implies | |||
sin y | = | x. | |
Therefore, on taking the derivative with respect to x: | |||
= | 1; | ||
= | |||
= |
according to Problem 1.
That is what we wanted to prove.
Note: We could have used the theorem of Lesson 8 directly:
We will use that theorem in the proofs that follow.
Problem 3. Calculate these derivatives. [In parts a) and b), use the chain rule.]
a) | d dx |
arcsin x2 | = |
b) | d dx |
= |
c) | d dx |
x2 arcsin x | = |
The derivative of y = arccos x
The derivative of arccos x is the negative of the derivative
of arcsin x. That will be true for the inverse of each pair of cofunctions.
The derivative of arccot x will be the negative
of the derivative of arctan x.
The derivative of arccsc x will be the negative
of the derivative of arcsec x.
For, beginning with arccos x:
The angle whose cosine is x is the complement
of the angle whose sine is x.
arccos x | = | π 2 |
− arcsin x. |
Since the derivative of | π 2 |
is 0, the result follows. |
Problem 4. Calculate these derivatives.
a) | d dx |
arccos | x a |
= |
b) | d dx |
x arccos 2x | = |
The derivative of y = arctan x
d dx |
arctan x | = | 1 1 + x2 |
First,
y = arctan x implies tan y = x.
Therefore, according to the theorem of Lesson 8:
Lesson 12. | ||
, according to the Pythagorean identity b, |
||
Which is what we wanted to prove.
Therefore, the derivative of arccot x is its negative:
d dx |
arccot x | = − | 1 1 + x2 |
Problem 5. Calculate these derivatives.
a) | d dx |
arctan (ax2) | = | 2ax 1 + a2x4 |
b) | d dx |
arccot | x a |
= | −a a2 + x2 |
c) | d dx |
arctan | 2 x |
= | −2 x2 + 4 |
d) | d dx |
arccot 2x | = | −2 4x2 + 1 |
*
The remaining derivatives come up rarely in calculus. Nevertheless, here are the proofs.
The derivative of y = arcsec x
Again,
y = arcsec x implies sec y = x.
Therefore, according to the theorem of Lesson 8:
Now, according to the theorem of Topic 19 of Trigonometry: that product is never negative. Therefore to ensure that, rather than replacing sec y with x, we will replace it with |x|. And in Problem 2 we will take only the positive root of tan y.
Therefore,
Which is what we wanted to prove.
If we took the range of arcsec x to be a third quadrant angle between −π and −π/2, when x is negative, then we would not need to write the absolute value, and the proof would be straightforward. We would simply replace sec y with x, and take the positive root of tan y, because tan y is positive in the first and third quadrants. In the graph of y = arcsec x with that range, the slope for negative x is negative. The disadvantage of taking that range is that, when x is negative, arcsec x will not equal arccos 1/x, because arccos 1/x will be a 2nd quadrant angle. But then in the proof we have to write the absolute value.
The derivative, therefore, of arccsc x is its negative:
d dx |
arccsc x | = − |
Next Lesson: Derivatives of exponential and logarithmic functions
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