Differentiation of trigonometric functions

Trigonometry

• Outline
• History
• Usage

• Functions (sin, cos, tan, inverse)
• Generalized trigonometry

Reference

• Identities
• Exact constants
• Tables
• Unit circle

Laws and theorems

• Sines
• Cosines
  • Pythagorean theorem
• Tangents
• Cotangents

Calculus

• Trigonometric substitution
• Integrals (inverse functions)
• Derivatives
• Trigonometric series

Mathematicians

• Hipparchus
• Ptolemy
• Brahmagupta
• al-Hasib
• al-Battani
• Regiomontanus
• Viète
• de Moivre
• Euler
• Fourier

Function Derivative sin ⁡ ( x ) {displaystyle sin(x)} cos ⁡ ( x ) {displaystyle cos(x)} cos ⁡ ( x ) {displaystyle cos(x)} − sin ⁡ ( x ) {displaystyle -sin(x)} tan ⁡ ( x ) {displaystyle tan(x)} sec 2 ⁡ ( x ) {displaystyle sec ^{2}(x)} cot ⁡ ( x ) {displaystyle cot(x)} − csc 2 ⁡ ( x ) {displaystyle -csc ^{2}(x)} sec ⁡ ( x ) {displaystyle sec(x)} sec ⁡ ( x ) tan ⁡ ( x ) {displaystyle sec(x)tan(x)} csc ⁡ ( x ) {displaystyle csc(x)} − csc ⁡ ( x ) cot ⁡ ( x ) {displaystyle -csc(x)cot(x)} arcsin ⁡ ( x ) {displaystyle arcsin(x)} 1 1 − x 2 {displaystyle {frac {1}{sqrt {1-x^{2}}}}} arccos ⁡ ( x ) {displaystyle arccos(x)} − 1 1 − x 2 {displaystyle -{frac {1}{sqrt {1-x^{2}}}}} arctan ⁡ ( x ) {displaystyle arctan(x)} 1 x 2 + 1 {displaystyle {frac {1}{x^{2}+1}}} arccot ⁡ ( x ) {displaystyle operatorname {arccot}(x)} − 1 x 2 + 1 {displaystyle -{frac {1}{x^{2}+1}}} arcsec ⁡ ( x ) {displaystyle operatorname {arcsec}(x)} 1 | x | x 2 − 1 {displaystyle {frac {1}{|x|{sqrt {x^{2}-1}}}}} arccsc ⁡ ( x ) {displaystyle operatorname {arccsc}(x)} − 1 | x | x 2 − 1 {displaystyle -{frac {1}{|x|{sqrt {x^{2}-1}}}}}

The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. For example, the derivative of the sine function is written sin′(a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle.

All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule applied to functions such as tan(x) = sin(x)/cos(x). Knowing these derivatives, the derivatives of the inverse trigonometric functions are found using implicit differentiation.

The diagram at right shows a circle with centre O and radius r = 1. Let two radii OA and OB make an arc of θ radians. Since we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < ⁠1/2⁠ π in the first quadrant.

In the diagram, let R1 be the triangle OAB, R2 the circular sector OAB, and R3 the triangle OAC.

The area of triangle OAB is:

A r e a ( R 1 ) = 1 2 | O A | | O B | sin ⁡ θ = 1 2 sin ⁡ θ . {displaystyle mathrm {Area} (R_{1})={tfrac {1}{2}} |OA| |OB|sin theta ={tfrac {1}{2}}sin theta ,.}

The area of the circular sector OAB is:

A r e a ( R 2 ) = 1 2 θ . {displaystyle mathrm {Area} (R_{2})={tfrac {1}{2}}theta ,.}

The area of the triangle OAC is given by:

A r e a ( R 3 ) = 1 2 | O A | | A C | = 1 2 tan ⁡ θ . {displaystyle mathrm {Area} (R_{3})={tfrac {1}{2}} |OA| |AC|={tfrac {1}{2}}tan theta ,.}

Since each region is contained in the next, one has:

Area ( R 1 ) < Area ( R 2 ) < Area ( R 3 ) ⟹ 1 2 sin ⁡ θ < 1 2 θ < 1 2 tan ⁡ θ . {displaystyle {text{Area}}(R_{1})<{text{Area}}(R_{2})<{text{Area}}(R_{3})implies {tfrac {1}{2}}sin theta <{tfrac {1}{2}}theta <{tfrac {1}{2}}tan theta ,.}

Moreover, since sin θ > 0 in the first quadrant, we may divide through by ⁠1/2⁠ sin θ, giving: 1 < θ sin ⁡ θ < 1 cos ⁡ θ ⟹ 1 > sin ⁡ θ θ > cos ⁡ θ . {displaystyle 1<{frac {theta }{sin theta }}<{frac {1}{cos theta }}implies 1>{frac {sin theta }{theta }}>cos theta ,.}

In the last step we took the reciprocals of the three positive terms, reversing the inequities.

We conclude that for 0 < θ < ⁠1/2⁠ π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ). Thus, as θ gets closer to 0, sin(θ)/θ is "squeezed" between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/θ must tend to 1 as θ tends to 0 from the positive side:

lim θ → 0 + sin ⁡ θ θ = 1 . {displaystyle lim _{theta to 0^{+}}{frac {sin theta }{theta }}=1,.} [1]

For the case where θ is a small negative number -⁠1/2⁠ π < θ < 0, we use the fact that sine is an odd function:

lim θ → 0 − sin ⁡ θ θ = lim θ → 0 + sin ⁡ ( − θ ) − θ = lim θ → 0 + − sin ⁡ θ − θ = lim θ → 0 + sin ⁡ θ θ = 1 . {displaystyle lim _{theta to 0^{-}}!{frac {sin theta }{theta }} = lim _{theta to 0^{+}}!{frac {sin(-theta )}{-theta }} = lim _{theta to 0^{+}}!{frac {-sin theta }{-theta }} = lim _{theta to 0^{+}}!{frac {sin theta }{theta }} = 1,.}

The last section enables us to calculate this new limit relatively easily. This is done by employing a simple trick. In this calculation, the sign of θ is unimportant.

lim θ → 0 cos ⁡ θ − 1 θ = lim θ → 0 ( cos ⁡ θ − 1 θ ) ( cos ⁡ θ + 1 cos ⁡ θ + 1 ) = lim θ → 0 cos 2 θ − 1 θ ( cos ⁡ θ + 1 ) . {displaystyle lim _{theta to 0},{frac {cos theta -1}{theta }} = lim _{theta to 0}left({frac {cos theta -1}{theta }}right)!!left({frac {cos theta +1}{cos theta +1}}right) = lim _{theta to 0},{frac {cos ^{2}!theta -1}{theta ,(cos theta +1)}}.}

Using cos2θ - 1 = -sin2θ, the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:

lim θ → 0 cos ⁡ θ − 1 θ = lim θ → 0 − sin 2 ⁡ θ θ ( cos ⁡ θ + 1 ) = ( − lim θ → 0 sin ⁡ θ θ ) ( lim θ → 0 sin ⁡ θ cos ⁡ θ + 1 ) = ( − 1 ) ( 0 2 ) = 0 . {displaystyle lim _{theta to 0},{frac {cos theta -1}{theta }} = lim _{theta to 0},{frac {-sin ^{2}theta }{theta (cos theta +1)}} = left(-lim _{theta to 0}{frac {sin theta }{theta }}right)!left(lim _{theta to 0},{frac {sin theta }{cos theta +1}}right) = (-1)left({frac {0}{2}}right)=0,.}

Using the limit for the sine function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find:

lim θ → 0 tan ⁡ θ θ = ( lim θ → 0 sin ⁡ θ θ ) ( lim θ → 0 1 cos ⁡ θ ) = ( 1 ) ( 1 ) = 1 . {displaystyle lim _{theta to 0}{frac {tan theta }{theta }} = left(lim _{theta to 0}{frac {sin theta }{theta }}right)!left(lim _{theta to 0}{frac {1}{cos theta }}right) = (1)(1) = 1,.}

We calculate the derivative of the sine function from the limit definition:

d d θ sin ⁡ θ = lim δ → 0 sin ⁡ ( θ + δ ) − sin ⁡ θ δ . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},sin theta =lim _{delta to 0}{frac {sin(theta +delta )-sin theta }{delta }}.}

Using the angle addition formula sin(α+β) = sin α cos β + sin β cos α, we have:

d d θ sin ⁡ θ = lim δ → 0 sin ⁡ θ cos ⁡ δ + sin ⁡ δ cos ⁡ θ − sin ⁡ θ δ = lim δ → 0 ( sin ⁡ δ δ cos ⁡ θ + cos ⁡ δ − 1 δ sin ⁡ θ ) . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},sin theta =lim _{delta to 0}{frac {sin theta cos delta +sin delta cos theta -sin theta }{delta }}=lim _{delta to 0}left({frac {sin delta }{delta }}cos theta +{frac {cos delta -1}{delta }}sin theta right).}

Using the limits for the sine and for the cosine functions:

d d θ sin ⁡ θ = ( 1 ) cos ⁡ θ + ( 0 ) sin ⁡ θ = cos ⁡ θ . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},sin theta =(1)cos theta +(0)sin theta =cos theta ,.}

We again calculate the derivative of the cosine function from the limit definition:

d d θ cos ⁡ θ = lim δ → 0 cos ⁡ ( θ + δ ) − cos ⁡ θ δ . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},cos theta =lim _{delta to 0}{frac {cos(theta +delta )-cos theta }{delta }}.}

Using the angle addition formula cos(α+β) = cos α cos β - sin α sin β, we have:

d d θ cos ⁡ θ = lim δ → 0 cos ⁡ θ cos ⁡ δ − sin ⁡ θ sin ⁡ δ − cos ⁡ θ δ = lim δ → 0 ( cos ⁡ δ − 1 δ cos ⁡ θ − sin ⁡ δ δ sin ⁡ θ ) . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},cos theta =lim _{delta to 0}{frac {cos theta cos delta -sin theta sin delta -cos theta }{delta }}=lim _{delta to 0}left({frac {cos delta -1}{delta }}cos theta ,-,{frac {sin delta }{delta }}sin theta right).}

Using the limits for the sine and cosine functions:

d d θ cos ⁡ θ = ( 0 ) cos ⁡ θ − ( 1 ) sin ⁡ θ = − sin ⁡ θ . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},cos theta =(0)cos theta -(1)sin theta =-sin theta ,.}

To compute the derivative of the cosine function from the chain rule, first observe the following three facts:

cos ⁡ θ = sin ⁡ ( π 2 − θ ) {displaystyle cos theta =sin left({tfrac {pi }{2}}-theta right)} sin ⁡ θ = cos ⁡ ( π 2 − θ ) {displaystyle sin theta =cos left({tfrac {pi }{2}}-theta right)} d d θ sin ⁡ θ = cos ⁡ θ {displaystyle {tfrac {operatorname {d} }{operatorname {d} !theta }}sin theta =cos theta }

The first and the second are trigonometric identities, and the third is proven above. Using these three facts, we can write the following,

d d θ cos ⁡ θ = d d θ sin ⁡ ( π 2 − θ ) {displaystyle {tfrac {operatorname {d} }{operatorname {d} !theta }}cos theta ={tfrac {operatorname {d} }{operatorname {d} !theta }}sin left({tfrac {pi }{2}}-theta right)}

We can differentiate this using the chain rule. Letting f ( x ) = sin ⁡ x , g ( θ ) = π 2 − θ {displaystyle f(x)=sin x, g(theta )={tfrac {pi }{2}}-theta } , we have:

d d θ f ( g ( θ ) ) = f ′ ( g ( θ ) ) ⋅ g ′ ( θ ) = cos ⁡ ( π 2 − θ ) ⋅ ( 0 − 1 ) = − sin ⁡ θ {displaystyle {tfrac {operatorname {d} }{operatorname {d} !theta }}f!left(g!left(theta right)right)=f^{prime }!left(g!left(theta right)right)cdot g^{prime }!left(theta right)=cos left({tfrac {pi }{2}}-theta right)cdot (0-1)=-sin theta } .

Therefore, we have proven that

d d θ cos ⁡ θ = − sin ⁡ θ {displaystyle {tfrac {operatorname {d} }{operatorname {d} !theta }}cos theta =-sin theta } .

To calculate the derivative of the tangent function tan θ, we use first principles. By definition:

d d θ tan ⁡ θ = lim δ → 0 ( tan ⁡ ( θ + δ ) − tan ⁡ θ δ ) . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},tan theta =lim _{delta to 0}left({frac {tan(theta +delta )-tan theta }{delta }}right).}

Using the well-known angle formula tan(α+β) = (tan α + tan β) / (1 - tan α tan β), we have:

d d θ tan ⁡ θ = lim δ → 0 [ tan ⁡ θ + tan ⁡ δ 1 − tan ⁡ θ tan ⁡ δ − tan ⁡ θ δ ] = lim δ → 0 [ tan ⁡ θ + tan ⁡ δ − tan ⁡ θ + tan 2 ⁡ θ tan ⁡ δ δ ( 1 − tan ⁡ θ tan ⁡ δ ) ] . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},tan theta =lim _{delta to 0}left[{frac {{frac {tan theta +tan delta }{1-tan theta tan delta }}-tan theta }{delta }}right]=lim _{delta to 0}left[{frac {tan theta +tan delta -tan theta +tan ^{2}theta tan delta }{delta left(1-tan theta tan delta right)}}right].}

Using the fact that the limit of a product is the product of the limits:

d d θ tan ⁡ θ = lim δ → 0 tan ⁡ δ δ × lim δ → 0 ( 1 + tan 2 ⁡ θ 1 − tan ⁡ θ tan ⁡ δ ) . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},tan theta =lim _{delta to 0}{frac {tan delta }{delta }}times lim _{delta to 0}left({frac {1+tan ^{2}theta }{1-tan theta tan delta }}right).}

Using the limit for the tangent function, and the fact that tan δ tends to 0 as δ tends to 0:

d d θ tan ⁡ θ = 1 × 1 + tan 2 ⁡ θ 1 − 0 = 1 + tan 2 ⁡ θ . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},tan theta =1times {frac {1+tan ^{2}theta }{1-0}}=1+tan ^{2}theta .}

We see immediately that:

d d θ tan ⁡ θ = 1 + sin 2 ⁡ θ cos 2 ⁡ θ = cos 2 ⁡ θ + sin 2 ⁡ θ cos 2 ⁡ θ = 1 cos 2 ⁡ θ = sec 2 ⁡ θ . {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }},tan theta =1+{frac {sin ^{2}theta }{cos ^{2}theta }}={frac {cos ^{2}theta +sin ^{2}theta }{cos ^{2}theta }}={frac {1}{cos ^{2}theta }}=sec ^{2}theta ,.}

One can also compute the derivative of the tangent function using the quotient rule.

d d θ tan ⁡ θ = d d θ sin ⁡ θ cos ⁡ θ = ( sin ⁡ θ ) ′ ⋅ cos ⁡ θ − sin ⁡ θ ⋅ ( cos ⁡ θ ) ′ cos 2 ⁡ θ = cos 2 ⁡ θ + sin 2 ⁡ θ cos 2 ⁡ θ {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }}tan theta ={frac {operatorname {d} }{operatorname {d} !theta }}{frac {sin theta }{cos theta }}={frac {left(sin theta right)^{prime }cdot cos theta -sin theta cdot left(cos theta right)^{prime }}{cos ^{2}theta }}={frac {cos ^{2}theta +sin ^{2}theta }{cos ^{2}theta }}}

The numerator can be simplified to 1 by the Pythagorean identity, giving us,

1 cos 2 ⁡ θ = sec 2 ⁡ θ {displaystyle {frac {1}{cos ^{2}theta }}=sec ^{2}theta }

Therefore,

d d θ tan ⁡ θ = sec 2 ⁡ θ {displaystyle {frac {operatorname {d} }{operatorname {d} !theta }}tan theta =sec ^{2}theta }

The following derivatives are found by setting a variable y equal to the inverse trigonometric function that we wish to take the derivative of. Using implicit differentiation and then solving for dy/dx, the derivative of the inverse function is found in terms of y. To convert dy/dx back into being in terms of x, we can draw a reference triangle on the unit circle, letting θ be y. Using the Pythagorean theorem and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x.

We let

y = arcsin ⁡ x {displaystyle y=arcsin x,!}

Where

− π 2 ≤ y ≤ π 2 {displaystyle -{frac {pi }{2}}leq yleq {frac {pi }{2}}}

Then

sin ⁡ y = x {displaystyle sin y=x,!}

Taking the derivative with respect to x {displaystyle x} on both sides and solving for dy/dx:

d d x sin ⁡ y = d d x x {displaystyle {d over dx}sin y={d over dx}x} cos ⁡ y ⋅ d y d x = 1 {displaystyle cos ycdot {dy over dx}=1,!}

Substituting cos ⁡ y = 1 − sin 2 ⁡ y {displaystyle cos y={sqrt {1-sin ^{2}y}}} in from above,

1 − sin 2 ⁡ y ⋅ d y d x = 1 {displaystyle {sqrt {1-sin ^{2}y}}cdot {dy over dx}=1}

Substituting x = sin ⁡ y {displaystyle x=sin y} in from above,

1 − x 2 ⋅ d y d x = 1 {displaystyle {sqrt {1-x^{2}}}cdot {dy over dx}=1} d y d x = 1 1 − x 2 {displaystyle {dy over dx}={frac {1}{sqrt {1-x^{2}}}}}

We let

y = arccos ⁡ x {displaystyle y=arccos x,!}

Where

0 ≤ y ≤ π {displaystyle 0leq yleq pi }

Then

cos ⁡ y = x {displaystyle cos y=x,!}

Taking the derivative with respect to x {displaystyle x} on both sides and solving for dy/dx:

d d x cos ⁡ y = d d x x {displaystyle {d over dx}cos y={d over dx}x} − sin ⁡ y ⋅ d y d x = 1 {displaystyle -sin ycdot {dy over dx}=1}

Substituting sin ⁡ y = 1 − cos 2 ⁡ y {displaystyle sin y={sqrt {1-cos ^{2}y}},!} in from above, we get

− 1 − cos 2 ⁡ y ⋅ d y d x = 1 {displaystyle -{sqrt {1-cos ^{2}y}}cdot {dy over dx}=1}

Substituting x = cos ⁡ y {displaystyle x=cos y,!} in from above, we get

− 1 − x 2 ⋅ d y d x = 1 {displaystyle -{sqrt {1-x^{2}}}cdot {dy over dx}=1} d y d x = − 1 1 − x 2 {displaystyle {dy over dx}=-{frac {1}{sqrt {1-x^{2}}}}}

Alternatively, once the derivative of arcsin ⁡ x {displaystyle arcsin x} is established, the derivative of arccos ⁡ x {displaystyle arccos x} follows immediately by differentiating the identity arcsin ⁡ x + arccos ⁡ x = π / 2 {displaystyle arcsin x+arccos x=pi /2} so that ( arccos ⁡ x ) ′ = − ( arcsin ⁡ x ) ′ {displaystyle (arccos x)'=-(arcsin x)'} .

We let

y = arctan ⁡ x {displaystyle y=arctan x,!}

Where

− π 2 < y < π 2 {displaystyle -{frac {pi }{2}}<y<{frac {pi }{2}}}

Then

tan ⁡ y = x {displaystyle tan y=x,!}

Taking the derivative with respect to x {displaystyle x} on both sides and solving for dy/dx:

d d x tan ⁡ y = d d x x {displaystyle {d over dx}tan y={d over dx}x}

Left side:

d d x tan ⁡ y = sec 2 ⁡ y ⋅ d y d x = ( 1 + tan 2 ⁡ y ) d y d x {displaystyle {d over dx}tan y=sec ^{2}ycdot {dy over dx}=(1+tan ^{2}y){dy over dx}} using the Pythagorean identity

Right side:

d d x x = 1 {displaystyle {d over dx}x=1}

Therefore,

( 1 + tan 2 ⁡ y ) d y d x = 1 {displaystyle (1+tan ^{2}y){dy over dx}=1}

Substituting x = tan ⁡ y {displaystyle x=tan y,!} in from above, we get

( 1 + x 2 ) d y d x = 1 {displaystyle (1+x^{2}){dy over dx}=1} d y d x = 1 1 + x 2 {displaystyle {dy over dx}={frac {1}{1+x^{2}}}}

We let

y = arccot ⁡ x {displaystyle y=operatorname {arccot} x}

where 0 < y < π {displaystyle 0<y<pi } . Then

cot ⁡ y = x {displaystyle cot y=x}

Taking the derivative with respect to x {displaystyle x} on both sides and solving for dy/dx:

d d x cot ⁡ y = d d x x {displaystyle {frac {d}{dx}}cot y={frac {d}{dx}}x}

Left side:

d d x cot ⁡ y = − csc 2 ⁡ y ⋅ d y d x = − ( 1 + cot 2 ⁡ y ) d y d x {displaystyle {d over dx}cot y=-csc ^{2}ycdot {dy over dx}=-(1+cot ^{2}y){dy over dx}} using the Pythagorean identity

Right side:

d d x x = 1 {displaystyle {d over dx}x=1}

Therefore,

− ( 1 + cot 2 ⁡ y ) d y d x = 1 {displaystyle -(1+cot ^{2}y){frac {dy}{dx}}=1}

Substituting x = cot ⁡ y {displaystyle x=cot y} ,

− ( 1 + x 2 ) d y d x = 1 {displaystyle -(1+x^{2}){frac {dy}{dx}}=1} d y d x = − 1 1 + x 2 {displaystyle {frac {dy}{dx}}=-{frac {1}{1+x^{2}}}}

Alternatively, as the derivative of arctan ⁡ x {displaystyle arctan x} is derived as shown above, then using the identity arctan ⁡ x + arccot ⁡ x = π 2 {displaystyle arctan x+operatorname {arccot} x={dfrac {pi }{2}}} follows immediately that d d x arccot ⁡ x = d d x ( π 2 − arctan ⁡ x ) = − 1 1 + x 2 {displaystyle {begin{aligned}{dfrac {d}{dx}}operatorname {arccot} x&={dfrac {d}{dx}}left({dfrac {pi }{2}}-arctan xright)&=-{dfrac {1}{1+x^{2}}}end{aligned}}}

Let

y = arcsec ⁡ x ∣ | x | ≥ 1 {displaystyle y=operatorname {arcsec} x mid |x|geq 1}

Then

x = sec ⁡ y ∣ y ∈ [ 0 , π 2 ) ∪ ( π 2 , π ] {displaystyle x=sec ymid yin left[0,{frac {pi }{2}}right)cup left({frac {pi }{2}},pi right]} d x d y = sec ⁡ y tan ⁡ y = | x | x 2 − 1 {displaystyle {frac {dx}{dy}}=sec ytan y=|x|{sqrt {x^{2}-1}}}

(The absolute value in the expression is necessary as the product of secant and tangent in the interval of y is always nonnegative, while the radical x 2 − 1 {displaystyle {sqrt {x^{2}-1}}} is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)

d y d x = 1 | x | x 2 − 1 {displaystyle {frac {dy}{dx}}={frac {1}{|x|{sqrt {x^{2}-1}}}}}

Alternatively, the derivative of arcsecant may be derived from the derivative of arccosine using the chain rule.

Let

y = arcsec ⁡ x = arccos ⁡ ( 1 x ) {displaystyle y=operatorname {arcsec} x=arccos left({frac {1}{x}}right)}

Where

| x | ≥ 1 {displaystyle |x|geq 1} and y ∈ [ 0 , π 2 ) ∪ ( π 2 , π ] {displaystyle yin left[0,{frac {pi }{2}}right)cup left({frac {pi }{2}},pi right]}

Then, applying the chain rule to arccos ⁡ ( 1 x ) {displaystyle arccos left({frac {1}{x}}right)} :

d y d x = − 1 1 − ( 1 x ) 2 ⋅ ( − 1 x 2 ) = 1 x 2 1 − 1 x 2 = 1 x 2 x 2 − 1 x 2 = 1 x 2 x 2 − 1 = 1 | x | x 2 − 1 {displaystyle {frac {dy}{dx}}=-{frac {1}{sqrt {1-({frac {1}{x}})^{2}}}}cdot left(-{frac {1}{x^{2}}}right)={frac {1}{x^{2}{sqrt {1-{frac {1}{x^{2}}}}}}}={frac {1}{x^{2}{frac {sqrt {x^{2}-1}}{sqrt {x^{2}}}}}}={frac {1}{{sqrt {x^{2}}}{sqrt {x^{2}-1}}}}={frac {1}{|x|{sqrt {x^{2}-1}}}}}

Let

y = arccsc ⁡ x ∣ | x | ≥ 1 {displaystyle y=operatorname {arccsc} x mid |x|geq 1}

Then

x = csc ⁡ y ∣ y ∈ [ − π 2 , 0 ) ∪ ( 0 , π 2 ] {displaystyle x=csc y mid yin left[-{frac {pi }{2}},0right)cup left(0,{frac {pi }{2}}right]} d x d y = − csc ⁡ y cot ⁡ y = − | x | x 2 − 1 {displaystyle {frac {dx}{dy}}=-csc ycot y=-|x|{sqrt {x^{2}-1}}}

(The absolute value in the expression is necessary as the product of cosecant and cotangent in the interval of y is always nonnegative, while the radical x 2 − 1 {displaystyle {sqrt {x^{2}-1}}} is always nonnegative by definition of the principal square root, so the remaining factor must also be nonnegative, which is achieved by using the absolute value of x.)

d y d x = − 1 | x | x 2 − 1 {displaystyle {frac {dy}{dx}}={frac {-1}{|x|{sqrt {x^{2}-1}}}}}

Alternatively, the derivative of arccosecant may be derived from the derivative of arcsine using the chain rule.

Let

y = arccsc ⁡ x = arcsin ⁡ ( 1 x ) {displaystyle y=operatorname {arccsc} x=arcsin left({frac {1}{x}}right)}

Where

| x | ≥ 1 {displaystyle |x|geq 1} and y ∈ [ − π 2 , 0 ) ∪ ( 0 , π 2 ] {displaystyle yin left[-{frac {pi }{2}},0right)cup left(0,{frac {pi }{2}}right]}

Then, applying the chain rule to arcsin ⁡ ( 1 x ) {displaystyle arcsin left({frac {1}{x}}right)} :

d y d x = 1 1 − ( 1 x ) 2 ⋅ ( − 1 x 2 ) = − 1 x 2 1 − 1 x 2 = − 1 x 2 x 2 − 1 x 2 = − 1 x 2 x 2 − 1 = − 1 | x | x 2 − 1 {displaystyle {frac {dy}{dx}}={frac {1}{sqrt {1-({frac {1}{x}})^{2}}}}cdot left(-{frac {1}{x^{2}}}right)=-{frac {1}{x^{2}{sqrt {1-{frac {1}{x^{2}}}}}}}=-{frac {1}{x^{2}{frac {sqrt {x^{2}-1}}{sqrt {x^{2}}}}}}=-{frac {1}{{sqrt {x^{2}}}{sqrt {x^{2}-1}}}}=-{frac {1}{|x|{sqrt {x^{2}-1}}}}}

• Calculus - Branch of mathematics
• Derivative - Instantaneous rate of change (mathematics)
• Differentiation rules - Rules for computing derivatives of functions
• General Leibniz rule - Generalization of the product rule in calculus
• Inverse functions and differentiation - Formula for the derivative of an inverse functionPages displaying short descriptions of redirect targets
• Linearity of differentiation - Calculus property
• List of integrals of inverse trigonometric functions
• List of trigonometric identities
• Trigonometry - Area of geometry, about angles and lengths

• Handbook of Mathematical Functions, Edited by Abramowitz and Stegun, National Bureau of Standards, Applied Mathematics Series, 55 (1964)