Mathematical process of finding the derivative of a trigonometric function
Thedifferentiation of trigonometric functions is the mathematical process of finding thederivative of atrigonometric 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 anglex = 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 thequotient rule applied to functions such as tan(x ) = sin(x )/cos(x ). Knowing these derivatives, the derivatives of theinverse trigonometric functions are found usingimplicit differentiation .
Proofs of derivatives of trigonometric functions [ edit ] [ edit ] Circle, centreO , radius 1
The diagram at right shows a circle with centreO and radiusr = 1. Let two radiiOA andOB make an arc of θ radians. Since we are considering the limit asθ tends to zero, we may assumeθ is a small positive number, say 0 < θ <1 / 2 π in the first quadrant.
In the diagram, letR 1 be the triangleOAB ,R 2 thecircular sector OAB , andR 3 the triangleOAC .
Thearea 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 \,.} Thearea 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 triangleOAC 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, sincesinθ > 0 in the first quadrant, we may divide through by1 / 2 sinθ , 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.
Squeeze: The curvesy = 1y = cosθ y = sin(θ )/θ
We conclude that for 0 < θ <1 / 2 π, the quantitysin(θ )/θ isalways less than 1 andalways greater than cos(θ). Thus, asθ gets closer to 0,sin(θ )/θ is "squeezed " between a ceiling at height 1 and a floor at heightcosθ , which rises towards 1; hence sin(θ )/θ must tend to 1 asθ tends to 0 from the positive side:
lim θ → 0 + sin θ θ = 1 . {\displaystyle \lim _{\theta \to 0^{+}}{\frac {\sin \theta }{\theta }}=1\,.}
For the case whereθ is a small negative number –1 / 2 π < θ < 0, we use the fact that sine is anodd 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\,.} [ edit ] 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 ofθ 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)}}.} Usingcos2 θ – 1 = –sin2 θ , 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\,.} [ edit ] Using the limit for thesine 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\,.} Derivative of the sine function [ edit ] We calculate the derivative of thesine function from thelimit 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 theangle 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 thesine andcosine 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 \,.} Derivative of the cosine function [ edit ] From the definition of derivative [ edit ] We again calculate the derivative of thecosine 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 formulacos(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 thesine andcosine 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 \,.} From the chain rule [ edit ] 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 aretrigonometric 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 thechain rule . Lettingf ( x ) = sin x , g ( θ ) = π 2 − θ {\displaystyle f(x)=\sin x,\ \ g(\theta )={\tfrac {\pi }{2}}-\theta }
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 } Derivative of the tangent function [ edit ] From the definition of derivative [ edit ] To calculate the derivative of thetangent function tanθ , we usefirst 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 formulatan(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 thetangent function, and the fact that tanδ 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 =1\times {\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 \,.} From the quotient rule [ edit ] One can also compute the derivative of the tangent function using thequotient 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 thePythagorean 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 } Proofs of derivatives of inverse trigonometric functions [ edit ] The following derivatives are found by setting avariable y equal to theinverse trigonometric function that we wish to take the derivative of. Usingimplicit differentiation and then solving fordy /dx , the derivative of the inverse function is found in terms ofy . To convertdy /dx back into being in terms ofx , we can draw a reference triangle on the unit circle, lettingθ be y. Using thePythagorean theorem and the definition of the regular trigonometric functions, we can finally expressdy /dx in terms ofx .
Differentiating the inverse sine function [ edit ] We let
y = arcsin x {\displaystyle y=\arcsin x\,\!} Where
− π 2 ≤ y ≤ π 2 {\displaystyle -{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}} Then
sin y = x {\displaystyle \sin y=x\,\!} Taking the derivative with respect tox {\displaystyle x}
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 y\cdot {dy \over dx}=1\,\!} Substitutingcos y = 1 − sin 2 y {\displaystyle \cos y={\sqrt {1-\sin ^{2}y}}}
1 − sin 2 y ⋅ d y d x = 1 {\displaystyle {\sqrt {1-\sin ^{2}y}}\cdot {dy \over dx}=1} Substitutingx = sin y {\displaystyle x=\sin y}
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}}}}} Differentiating the inverse cosine function [ edit ] We let
y = arccos x {\displaystyle y=\arccos x\,\!} Where
0 ≤ y ≤ π {\displaystyle 0\leq y\leq \pi } Then
cos y = x {\displaystyle \cos y=x\,\!} Taking the derivative with respect tox {\displaystyle x}
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 y\cdot {dy \over dx}=1} Substitutingsin y = 1 − cos 2 y {\displaystyle \sin y={\sqrt {1-\cos ^{2}y}}\,\!}
− 1 − cos 2 y ⋅ d y d x = 1 {\displaystyle -{\sqrt {1-\cos ^{2}y}}\cdot {dy \over dx}=1} Substitutingx = cos y {\displaystyle x=\cos y\,\!}
− 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 ofarcsin x {\displaystyle \arcsin x} arccos x {\displaystyle \arccos x} arcsin x + arccos x = π / 2 {\displaystyle \arcsin x+\arccos x=\pi /2} ( arccos x ) ′ = − ( arcsin x ) ′ {\displaystyle (\arccos x)'=-(\arcsin x)'}
Differentiating the inverse tangent function [ edit ] 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 tox {\displaystyle x}
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}y\cdot {dy \over dx}=(1+\tan ^{2}y){dy \over dx}} 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} Substitutingx = tan y {\displaystyle x=\tan y\,\!}
( 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}}}} Differentiating the inverse cotangent function [ edit ] We let
y = arccot x {\displaystyle y=\operatorname {arccot} x} where0 < y < π {\displaystyle 0<y<\pi }
cot y = x {\displaystyle \cot y=x} Taking the derivative with respect tox {\displaystyle x}
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}y\cdot {dy \over dx}=-(1+\cot ^{2}y){dy \over dx}} 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} Substitutingx = 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 ofarctan x {\displaystyle \arctan x} arctan x + arccot x = π 2 {\displaystyle \arctan x+\operatorname {arccot} x={\dfrac {\pi }{2}}} 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 x\right)\\&=-{\dfrac {1}{1+x^{2}}}\end{aligned}}}
Differentiating the inverse secant function [ edit ] Using implicit differentiation [ edit ] 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 y\mid \ y\in \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 y\tan 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 radicalx 2 − 1 {\displaystyle {\sqrt {x^{2}-1}}}
d y d x = 1 | x | x 2 − 1 {\displaystyle {\frac {dy}{dx}}={\frac {1}{|x|{\sqrt {x^{2}-1}}}}} Using the chain rule [ edit ] Alternatively, the derivative of arcsecant may be derived from the derivative of arccosine using thechain 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} y ∈ [ 0 , π 2 ) ∪ ( π 2 , π ] {\displaystyle y\in \left[0,{\frac {\pi }{2}}\right)\cup \left({\frac {\pi }{2}},\pi \right]} Then, applying the chain rule toarccos ( 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}}}}} Differentiating the inverse cosecant function [ edit ] Using implicit differentiation [ edit ] 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 \ y\in \left[-{\frac {\pi }{2}},0\right)\cup \left(0,{\frac {\pi }{2}}\right]} d x d y = − csc y cot y = − | x | x 2 − 1 {\displaystyle {\frac {dx}{dy}}=-\csc y\cot 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 radicalx 2 − 1 {\displaystyle {\sqrt {x^{2}-1}}}
d y d x = − 1 | x | x 2 − 1 {\displaystyle {\frac {dy}{dx}}={\frac {-1}{|x|{\sqrt {x^{2}-1}}}}} Using the chain rule [ edit ] Alternatively, the derivative of arccosecant may be derived from the derivative of arcsine using thechain 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} y ∈ [ − π 2 , 0 ) ∪ ( 0 , π 2 ] {\displaystyle y\in \left[-{\frac {\pi }{2}},0\right)\cup \left(0,{\frac {\pi }{2}}\right]} Then, applying the chain rule toarcsin ( 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}}}}}
