Derivatives of Inverse Trigonometric Functions

In this section, we will learn the derivative formulas of the inverse trigonometric functions with their proofs.

Now we calculate the derivative of sin1x (or arc sin x).

Derivative of sin1x

ddx(sin1x)=11x2(|x|<1)

Proof: Note that sin1x is defined when |x|1. We assume that
y=sin1x. As |x|1, we must have that π2yπ2. y=sin1xx=siny Differentiating with respect to y, we have dxdy=ddx(siny)=cosy dxdy=1sin2y dxdy=1x2 Therefore, we deduce that dydx=1dx/dy=11x2 So the derivative of sin1x is 11x2, where |x|1.

 

Next, we find the derivative of cos1x (or arc cos x).

Derivative of cos1x

ddx(cos1x)=11x2(|x|<1)

Proof: We know that sin1x+cos1x=π2 Differentiating with respect to x, we get ddx(sin1x+cos1x)=ddx(π2)ddx(sin1x)+ddx(cos1x)=0

[as the derivative of a constant is zero]

11x2+ddx(cos1x)=0 ddx(cos1x)=11x2 So the derivative of cos1x is 11x2, where |x|1.

 

Let us now calculate the derivative of tan1x (or arc tan x).

Derivative of tan1x

ddx(tan1x)=11+x2(<x<)

Proof: Note that tan1x is defined when <x<. We assume that y=tan1x. As <x<, we must have that π2yπ2. y=tan1xx=tany Differentiating with respect to y, we have dxdy=ddx(tany)=sec2y dxdy=1+tan2y dxdy=1+x2 Therefore, we deduce that dydx=1dx/dy=11+x2 So the derivative of tan1x is 11+x2, where <x<.

 

Next, we find the derivative of cot1x (or arc cot x).

Derivative of cot1x

ddx(cot1x)=11+x2(<x<)
Proof:  We know that tan1x+cot1x=π2 Differentiating with respect to x, we get ddx(tan1x+cot1x)=ddx(π2) ddx(tan1x)+ddx(cot1x)=0

[as the derivative of a constant is zero]

11+x2+ddx(cot1x)=0 ddx(cot1x)=11x2 So the derivative of cot1x is 11+x2, where <x<.

 

We now calculate the derivative of sec1x (or arc sec x).

Derivative of sec1x

ddx(sec1x)=1xx21(|x|>1)
Proof: Note that sec1x is defined when |x|>1. We assume that y=sec1x. x=secy Differentiating with respect to y, we have dxdy=ddx(secy)=secytany dxdy=secysec2y1 dxdy=xx21 Therefore, we deduce that dydx=1dx/dy=1xx21 So the derivative of sec1x is 1xx21, where |x|>1.

 

Next, we find the derivative of cosec1x (or arc cosec x).

Derivative of cosec1x

ddx(cosec1x)=1xx21(|x|>1)
Proof: We know that sec1x+cosec1x=π2 Differentiating with respect to x, we get ddx(sec1x+cosec1x)=ddx(π2) ddx(sec1x)+ddx(cosec1x)=0

 [as the derivative of a constant is zero]

1xx21+ddx(cosec1x)=0 ddx(cosec1x)=1xx21 So the derivative of cosec1x is 1xx21, where |x|>1.

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