Derivative of Inverse Trigonometric functions

In this section, we'll look at the derivatives of inverse trig functions. We'll need the formula from the previous section relating the derivatives of inverse functions to derive the derivatives of inverse trig functions. If f(x) and g(x) are inverse functions, then g'(x)=1f'(x)

Introduction

Arcus functions, cyclometric functions, and anti-trigonometric functions are all names for inverse trigonometric functions. These functions determine the angle for a given trigonometric value. In engineering, geometry, navigation, and other fields, inverse trigonometric functions are useful.

Inverse Sine

y=sin-1xsiny=x for -2y2

So evaluating an inverse trig function is equivalent to asking what angle (i.e. y) we plugged into the sine function to get x. The above restrictions on y are there to ensure that we get a consistent answer from the inverse sine. We know that there are an infinite number of possible angles, and we want a consistent value when working with inverse sine. Using the above range of angles yields all possible sine function values exactly once. If you're not sure, draw a unit circle and you'll see that that range of angles (the y's) will cover all possible sine values.

Note as well that since -1sin(y)1 we also have -1x1 

The inverse sine function and the sine function have the following relationship.

sinsin-1x=x sin-1sinx=x

In other words, they are polar opposites. This means that we can use the preceding fact to calculate the derivative of inverse sine. To begin with,

f(x)=sinx

g(x)=sin-1x

g'(x)=1f'(g(x))=1Cossin-1x

This is an ineffective formula. Let us see if we can come up with a better formula. To begin, recall the definition of the inverse sine function.

y=sin-1(x) x=siny

Using the first part of this definition, the derivative's denominator becomes,

Cossin-1x=cos(y)

Cos2y+Sin2y=1 Cos2y=1-Sin2y

Now apply the second part of the inverse sine function definition. The common denominator is then,

Cossin-1x=1-Sin2y=1-x2

When all of this is added up, the following derivative is obtained.

ddxsin-1x=11-x2

Inverse Cosine

y=Cos-1xCos y=x for 0y

We have a restriction on the angles, y, that we get from the inverse cosine function, just like we do with the inverse sine function. Again, if you want to double-check, a quick sketch of a unit circle should convince you that this range will cover all possible cosine values exactly once. In addition, we have,

-1x1 because -1Cos(y)1

Because the inverse cosine and cosine functions are inverses of each other, we have,

CosCos-1x=x Cos-1Cosx=x

We'll do the same thing we did with the inverse sine above to find the derivative. To begin with,

f(x)=Cosx

g(x)=Cos-1x

g'(x)=1f'(g(x))=1-sinCos-1x

Simplifying the denominator here is nearly identical to the work done for the inverse sine, so it isn't shown here. We get the following derivative after simplifying.

ddxCos-1x=11-x2

As a result, the inverse cosine derivative is nearly identical to the inverse sine derivative. The only distinction is the negative sign.

Inverse Tangent

y=tan-1xtan y=x for -2y2

Again, we have a constraint on y, but this time we can't let it go. Because tangent isn't defined at those two points, y can be either of the two endpoints in the restriction above. To ensure that this range covers all possible tangent values, draw a quick sketch of the tangent function. We can see that this range does indeed cover all possible tangent values. Furthermore, there are no restrictions on x in this case because tangent can take on any value.

We can ask for the limits of the inverse tangent function as x approaches plus or minus infinity because there is no restriction on x. We'll need the graph of the inverse tangent function to do this. This is depicted below.

From this graph we can see that

xtan-1x=2 xtan-1x=-2

Because the tangent and inverse tangent functions are inverse,

tantan-1x=x tan-1tanx=x

As a result, we can begin by calculating the derivative of the inverse tangent function.

f(x)=tanx

g(x)=tan-1x

We have,

y=tan-1x x=tan y

The denominator is then,

sec2tan-1x=sec2y

Now,

Cos2y+Sin2y=1

And divide every term by cos2y we will get,

1+tan2y=sec2y

The denominator is then,

sec2tan-1x=sec2y=1+tan2y

Finally, applying the second part of the definition of the inverse tangent function yields,

sec2tan-1x=1+tan2y=1+x2

The inverse tangent's derivative is then,

ddxtan-1x=11+x2

There are three more inverse trig functions, but these are the most common. The remaining three formulas could be derived in the same manner as the previous three. The derivatives of all six inverse trig functions are shown below.

• ddxsin-1x=11-x2
• ddxcos-1x=11-x2
• ddxtan-1x=11+x2
• ddxcot-1x=11+x2
• ddxsec-1x=1xx2-1
• ddxcosec-1x=1xx2-1

Alternate Notation

On occasion, an alternate notation is used to denote the inverse trig functions. This is the notation,

sin-1x=arcsin x cot-1x=arccot x

cos-1x=arccos x sec-1x=arcsec x

tan-1x=arctan x cosec-1x=arc cosec x

Conclusion

Inverse trigonometry is the reciprocal of trigonometric normal functions. The angle of the trigonometric functions is used to find the value of sides in trigonometry. In inverse trigonometry, on the other hand, the sides of the triangle are used to calculate the angle.