Find the Derivative - d/dx arctan(2x) Differentiate using the chain rule, which states that is where. Tap for more steps. To apply the Chain Rule, set as. Find the Derivative - d/dx y=arctan(1/x) Differentiate using the chain rule, which states that is where. Tap for more steps. To apply the Chain Rule, set as. The derivative of with respect to is. Replace all occurrences of with. Differentiate using.

1. Derivative Of Arctangent
2. Derivative Of Arctan 1
3. Derivative Of Arctan 3t

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13

IT IS NOT NECESSARY to memorize the derivatives of this Lesson. Rather, the student should know now to derive them.

In Topic 19 of Trigonometry, we introduced the inverse trigonometric functions. According to the inverse relations:

y = arcsin x implies sin y = x.

And similarly for each of the inverse trigonometric functions.

Problem 1. If y = arcsin x, show:

To see the answer, pass your mouse over the colored area.
To cover the answer again, click 'Refresh' ('Reload').
Do the problem yourself first!

Begin:

Derivative Of Arctangent

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:

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:

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.]

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.

Problem 4. Calculate these derivatives.

The derivative of y = arctan x

First,

y = arctan x implies tan y = x.

Therefore, according to the theorem of Lesson 8:

Which is what we wanted to prove.

Therefore, the derivative of arccot x is its negative:

Problem 5. Calculate these derivatives.

Derivative Of Arctan 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:

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