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 x^{2}  = 
b)  d dx 
= 
c)  d dx 
x^{2} 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 + x^{2} 
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 + x^{2} 
Problem 5. Calculate these derivatives.
a)  d dx 
arctan (ax^{2})  =  2ax 1 + a^{2}x^{4} 
b)  d dx 
arccot  x a 
=  −a a^{2} + x^{2} 
c)  d dx 
arctan  2 x 
=  −2 x^{2} + 4 
d)  d dx 
arccot 2x  =  −2 4x^{2} + 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
Copyright © 2017 Lawrence Spector
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