## Derivatives of Trigonometric Functions

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Derivs of Trig Functions: Basic Trig Functions :: Reciprocal Trig Functions :: Inverse Trig Functions

### DERIVATIVES OF BASIC TRIG FUNCTIONS:(top)

Important note: these derivatives are true only when the angle x is expressed in radians. This is because the limit rules we use to evaluate the limits hold only when x is expressed in degrees.

1. Derivative of Sine:

Explanation
We evaluate the derivative of sine using the definition of derivative:
 = by the definition of derivative = using the trigonometric identity for sin(x+y) = factoring out sin(x) from the terms containing it. = separating the limits using the sum rule and pulling the sin(x) and cos(x) terms outside the limits = by evaluating the trig limits; x must be in radians for these trig limits to hold = cos x simplifying the expression

2. Derivative of Cosine:

Explanation
We could evaluate the derivative of cosine from the definition of derivative, but it's much easier if we simply use some trig identities and the rule we just derived for derivative of sine:
 Since cos x = by the cofunction identities = , as we're doing the same thing to each side of the equation, = by the derivative of sine = by the cofunction identities
3. Derivative of Tangent:
Explanation
We can evaluate the derivative of tangent using the quotient rule and the derivatives for sine and cosine that we just developed:
 Since tan(x) = , (see quotient trig identities) = , since we've done the same thing to each side of the equation = by the quotient rule = by evaluating the derivatives for sin and cos = by simplifying = by the Pythagorean identity = sec2(x) by the definition of secant

### DERIVATIVES OF RECIPROCAL TRIG FUNCTIONS: (top)

To find the derivatives of the reciprocal trig functions, we'll simply use the quotient rule with their definitions in terms of the basic trig functions. Again, these derivatives are true only when the angle x is expressed in radians.

1. Derivative of Cosecant:

Explanation
 = by definition of cosecant = by the product rule = by evaluating the derivatives of 1 and sin = by simplifying and rearranging terms = by the definitions of cosecant and cotangent

2. Derivative of Secant:

Explanation
 = by the definition of secant = by the product rule = by evaluating the derivatives of 1 and cos = by simplifying and rearranging terms = by the definitions of secant and tangent

3. Derivative of Cotangent:

Explanation
 = by the definition of cotangent = by the product rule = by evaluating the derivatives of 1 and tan = by rewriting terms with identities = by rewriting fractional divison as multiplication = by canceling out the cos2(x) terms = –csc2(x) by the definition of cosecant

### DERIVATIVES OF INVERSE TRIG FUNCTIONS: (top)

Again, these derivatives are true only when the angle x is expressed in radians. We'll follow the same general strategy for calculating each of these derivatives. First, we'll rewrite the function to remove the inverse expression. We'll then differentiate implicitly, and we'll finish off by using trig to rewrite all of each derivative in terms of x.

1. Derivative of Arcsine:

Explanation
Let . We can get rid of the inverse trig function by rewriting this as .

Next, differentiate implicitly:

We must now replace "cos y" with some term involving x.
Since , the triangle at left is formed. The bottom leg is found using the Pythagorean theorem. Using this triangle, we can see that .

Substituting this into the equation for , we find that .

2. Derivative of Arccosine:
Explanation
Let . We can get rid of the inverse trig function by rewriting this as .

Next, differentiate implicitly:

We must now replace "cos y" with some term involving x.
Since , the triangle at left is formed. The bottom leg is found using the Pythagorean theorem. Using this triangle, we can see that .

Substituting this into the equation for , we find that .

3. Derivative of Arctangent:
Explanation
Let . We can get rid of the inverse trig function by rewriting this as .

Next, differentiate implicitly:

We must now replace "cos y" with some term involving x.
Since , the triangle at left is formed. The hypotenuse is found using the Pythagorean theorem. Using this triangle, we can see that .

Substituting this into the equation for , we find that .

Of course, each of the reciprocal trig functions—cosecant, secant, and cotangent—also has a corresponding inverse function. Here, we evaluate the derivatives of arccosecant, arcsecant, and arccotangent, using the same methods.

4. Derivative of Arccosecant:

Explanation
expl

5. Derivative of Arcsecant:

Explanation
Let . We can get rid of the inverse trig function by rewriting this as .

Next, differentiate implicitly:

We must now replace "(sec y)(tan y)" with some term involving x.
Since , the triangle at left is formed. The leg on the right side is found using the Pythagorean theorem. Using this triangle, we can see that .

Substituting this into the equation for , we find that .

6. Derivative of Arccotangent:

Explanation
expl

One final comment: If reciprocal ratios are inputs of inverse functions where the functions have reciprocal relationships, the resulting anges will be the same!

This fact can be used to rewrite problems and make them easier to solve.