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13

DERIVATIVES OF
INVERSE TRIGONOMETRIC FUNCTIONS

The derivative of y = arcsin x

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:

Inverse trigonometric functions

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:

  y = arcsin x
implies       
  1)   sin y = x.
 
  Therefore, according to the Pythagorean identity a':
 
  cos y = Inverse trigonometric functions
 
= Inverse trigonometric functions
  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:

Inverse trigonometric functions

Begin:

  y = arcsec x
implies       
  sec y = x.
 
  Therefore, according to the Pythagorean identity b:
 
  tan y = ±Inverse trigonometric functions
 
    = ±Inverse trigonometric functions

The derivative of y = arcsin x

Inverse trigonometric functions

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:
 
  Inverse trigonometric functions = 1;
  Inverse trigonometric functions = Inverse trigonometric functions
  Inverse trigonometric functions = Inverse trigonometric functions

according to Problem 1.

That is what we wanted to prove.

Note:  We could have used the theorem of Lesson 8 directly:

Inverse trigonometric functions

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 = Inverse trigonometric functions Inverse trigonometric functions
  b)    d 
dx
Inverse trigonometric functions = Inverse trigonometric functions
  c)    d 
dx
x2 arcsin x = Inverse trigonometric functions

The derivative of y = arccos x

Inverse trigonometric functions

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.

Inverse trigonometric functions

arccos x  =  π
2
 − arcsin x.
Since the derivative of  π
2
 is 0, the result follows.

Inverse trigonometric functions

Problem 4.   Calculate these derivatives.

  a)    d 
dx
arccos  x
a
 =  Inverse trigonometric functions
  b)    d 
dx
x arccos 2x  =  Inverse trigonometric functions

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:

Inverse trigonometric functions
 
  Inverse trigonometric functions Lesson 12.
 
  Inverse trigonometric functions , according to the Pythagorean
  identity b, 
 
  Inverse trigonometric functions

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

Inverse trigonometric functions

Again,

y = arcsec x   implies   sec y = x.

Therefore, according to the theorem of Lesson 8:

Inverse trigonometric functions
 
  Inverse trigonometric functions ,   Lesson 12.

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,

Inverse trigonometric functions

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  =  −  Inverse trigonometric functions

End of the lesson

Next Lesson:  Derivatives of exponential and logarithmic functions

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