INVERSE TR IGONOMETR ICFUNCT ION DER IVAT I VESmauricio bedoya1

July 28 2015

abstractIn this blog reader can find an easy method that show how to find thederivative of inverse trigonometric functions.

the problemWhile reading Schaum Integral and Derivative Calculus (Im getting aMathematical degree online), NO explanation is giving in the derivationof the formulas provided to find the derivatives of inverse trigonometricfunctions. I will show an easy method which requires the following knowl-edge:

1. Pythagoras Theorem.

2. Derivatives of trigonometric functions.

I will apply the method to: arcsin() and arctang().

arcsin() or Sin1 ()First make some assumption:

sin() = t (1)

Next, we draw (very important) the corresponding triangle according toPythagoras Theorem:

1

1 t2

t

Figure 1: Equation 1

1

2Next we find the derivative of equation (1):

cos()d = dt

d

dt=

1

cos()

Now go back to Figure 1 and replace cos():

d

dt=

11 t2

(2)

Equation 2 is the derivative of arcsin().

arctang() or T ang1 ()Just follow the same steps. First make some assumption:

Tang() = t (3)

Next, we draw (very important) the corresponding triangle according toPythagoras Theorem:

1+ t2

1

t

Figure 2: Equation 3Next we find the derivative of equation (3):

d

cos()2= dt

d

dt= cos()2

Now go back to Figure 2 and replace cos():

d

dt=

1

1+ t2(4)

Equation 4 is the derivative of arctang().

3applicationImagine that we have to find:

Sin1(2x 3) = y

If we follow the steps, then:

Sin(y) = 2x 3 (5)

Next, we draw (very important) the corresponding triangle according toPythagoras Theorem:

1

1 (2x 3)2

2x 3y

Figure 3: Equation 5Next we find the derivative of equation (5):

cos(y)dy = 2dx

dy

dx=

2

cos(y)

Now go back to Figure 3 and replace cos(y):

dy

dx=

21 (2x 3)2

(6)

Equation 6 is the desired result.

1 javierma36@hotmail.com