Derivative of inverse trigonometric functions

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    1. Review: what are "inverse trigonometric functions" ?

      Find the measure of angle ΞΈ\theta to the nearest degree:
      Derivative of inverse trigonometric functions
      1. Use implicit differentiation to prove the formula:
        ddx(sinβ‘βˆ’1x)=11βˆ’x2\frac{{d}}{{{d}x}}\left( {{{\sin }^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 - {x^2}} }}
        1. Calculate the derivative of y=y= arccos (x2)(x^2)
          1. Calculate the derivative of y=4y=4 arccot (3x+1)(3x+1)
            1. Calculate the derivative of y=4arctan⁑(ex)x2y= \frac{4 \arctan (e^x)}{x^2}
              1. Calculate the derivative of y=y= arcsecβ€…β€Šxβ€…β€Š\; x \; arccscβ€…β€Šx\; x
                1. Prove that ddx[tanβ‘βˆ’1(4x2)+cotβ‘βˆ’1(4x2)]=0\frac{d}{dx} [\tan^{-1}(4x^2)+ \cot^{-1}(4x^2)]=0
                  Topic Notes
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                  In this section, we will study the differential rules of inverse trigonometric functions, also known as cyclometric functions and arc-functions. Using our knowledge of inverse relations, and the definitions of the trigonometric functions "SOH CAH TOA", we will learn to derive the derivative formulas for inverse trig functions.

                  Derivative of Arctan

                  There are many students that find it easy to take derivatives of trig functions, but many struggle with derivatives of inverse trig functions. We will first talk about the many types of inverse trig functions we can differentiate, and then talk in detail about the first and second derivative of arctan.

                  Derivatives of inverse Trig Functions

                  First of all, there are exactly a total of 6 inverse trig functions. They are arcsin x, arccos x, arctan x, arcsec x, and arccsc x. However, some teachers use the power of -1 instead of arc to express them. For example, arcsin x is the same as sinβ‘βˆ’1x\sin^{-1} x. The derivative of each trig function is written below.

                  • Formula 1: Derivative of arcsinx
                    Formula 1: Derivative of arcsinx
                  • Formula 2: Derivative of arccosx
                    Formula 2: Derivative of arccosx
                  • Formula 3: Derivative of arctanx
                    Formula 3: Derivative of arctanx
                  • Formula 4: Derivative of arccotx
                    Formula 4: Derivative of arccotx
                  • Formula 5: Derivative of arcsecx
                    Formula 5: Derivative of arcsecx
                  • Formula 6: Derivative of arccscx
                    Formula 6: Derivative of arccscx

                  Notice that the derivative of arctan x and arccot x are very similar. In fact, they only differ by a negative sign so make sure you remember it correctly!

                  Derivative of arctan formula

                  Now a lot people can memorize these formulas, but not many people actually understand how we get them. For example, the derivative of arctan x is:

                  Figure 1: Derivative of arctan
                  Figure 1: Derivative of arctan

                  How is that even possible? Well first of all we want to start from the beginning. Let's take a look at:

                  • Equation 1: Derivative of arctan pt.1
                    Equation 1: Derivative of arctan pt.1
                  • Notice that is the same as y=tanβ‘βˆ’1xy = \tan^{-1} x.Now let us move the inverse tangent to the left side of the equation. Doing this will give us:

                    Equation 2: Derivative of arctan pt.2
                    Equation 2: Derivative of arctan pt.2
                  • Now what we want to do here is something called implicit differentiation. In other words, we want to take the derivative of both sides of the equation. Doing so will give us,

                    Equation 3: Derivative of arctan pt.3
                    Equation 3: Derivative of arctan pt.3

                    Notice that dydx\frac{dy}{dx} shows up in the equation because of the chain rule. If you don't understand chain rule or trig derivatives that well, then click those links!

                  • Now dividing both sides of the equation by sec⁑2y\sec^{2} y will give us:

                    Equation 4: Derivative of arctan pt.4
                    Equation 4: Derivative of arctan pt.4
                  • We are really close! Next, we look at the trig identity

                    Formula 7: Trig identity 1
                    Formula 7: Trig identity 1
                  • Notice that we can substitute this identity into our equation, which will give us:

                    Equation 5: Derivative of arctan pt.5
                    Equation 5: Derivative of arctan pt.5

                    Notice that this question requires you to know Pythagorean identities . So make sure you know them well before continuing with the question.

                  • Now going back on topic, recall that tan⁑y=x\tan y = x, so tan⁑2y=x2\tan^{2} y = x^{2}. Substituting this into the equation will finally give us the following formula:

                    Equation 6: Derivative of arctan pt.6
                    Equation 6: Derivative of arctan pt.6

                  If you are still having trouble understanding, I recommend you look at the intro video or look at this link, http://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/invTrigDeriv.html

                  Second derivative of arctan

                  First of all recall that the first derivative of arctan is:

                  • Equation 6: Derivative of arctan pt.6
                    Equation 6: Derivative of arctan pt.6
                  • Taking the second derivative is not that challenging because we only need to use the quotient rule. Recall that the formula of the quotient rule is:

                    Formula 8: Quotient Rule
                    Formula 8: Quotient Rule
                  • The second derivative will be:

                    Equation 7: Second Derivative of arctan pt.1
                    Equation 7: Second Derivative of arctan pt.1
                  • Hence we can conclude that:

                    Equation 8: Second Derivative of arctan pt.2
                    Equation 8: Second Derivative of arctan pt.2

                  Derivative of Other Inverse Trig Functions (arcsec)

                  What we start to notice after differentiating each inverse trig function is that they all have the same general idea. For example when taking the derivative of arctan, we moved the inverse tangent to the other side, did implicit differentiation, and used the trig identity. You will find that you can do the exact same thing for differentiating arccos x, arcsin x, and arccot x. The only difference is that the trig identity you use may be different. For example, arccos x and arcsin x uses the trig identity

                  • Formula 9: Trig Identity 2
                    Formula 9: Trig Identity 2
                  • However, the derivative of arcsec and arccsc is slightly more different. For example, let us try to take the derivative of arcsec. We start with the equation,

                    Equation 9: Derivative of arcsec pt.1
                    Equation 9: Derivative of arcsec pt.1
                  • Notice that y = arcsec x is the same as y=secβ‘βˆ’1xy = \sec^{-1} x. Now moving the inverse tangent to the other side of the equation will give us

                    Equation 10: Derivative of arcsec pt.2
                    Equation 10: Derivative of arcsec pt.2
                  • Using implicit differentiation again will give us the equation:

                    Equation 11: Derivative of arcsec pt.3
                    Equation 11: Derivative of arcsec pt.3
                  • Note that the derivative of sec⁑y\sec y is sec⁑ytan⁑y\sec y \tan y and the chain rule gives us dydx\frac{dy}{dx}. Now dividing both sides of the equation by sec⁑ytan⁑y\sec y \tan y gives us:

                    Equation 12: Derivative of arcsec pt.4
                    Equation 12: Derivative of arcsec pt.4
                  • Here is where it gets a little different.

                    Remember that we have the trig identity

                    Formula 7: Trig identity 1
                    Formula 7: Trig identity 1
                  • Now let us isolate tan in this trig identity. Doing a little bit of algebraic manipulation will give us that

                    Equation 13: Derivative of arcsec pt.5
                    Equation 13: Derivative of arcsec pt.5
                  • Now if we change all our ΞΈ \theta's to y's and substitute tan y in our derivative equation, then we will get that

                    Equation 14: Derivative of arcsec pt.6
                    Equation 14: Derivative of arcsec pt.6
                  • Lastly, we know that sec⁑y=x\sec y = x, hence substituting that will lead us to our equation:

                    Equation 15: Derivative of arcsec pt.7
                    Equation 15: Derivative of arcsec pt.7
                  • You may notice that our x in the equation is missing an absolute value. Why is that? It is because x can actually never be negative. If you want to know more about that in detail, please look at the last theorem in this link.

                    http://www.themathpage.com/aTrig/inverseTrig.htm#theo

                  • Hence we need to include the absolute value manually, which means that

                    Equation 16: Derivative of arcsec pt.8
                    Equation 16: Derivative of arcsec pt.8
                  Trigonometric Identities – "Pythagorean Identities"
                  sin2ΞΈβ€…β€Š+β€…β€Šcos2ΞΈβ€…β€Š=β€…β€Š1{si}{{n}^2}\theta \; + {\;co}{{s}^2}\theta \; = \;1
                  1β€…β€Š+β€…β€Štan2ΞΈβ€…β€Š=β€…β€Šsec2ΞΈ1{\;} + {\;ta}{{n}^2}\theta \; = \;{se}{{c}^2}\theta
                  1β€…β€Š+β€…β€Šcot2ΞΈβ€…β€Š=β€…β€Šcsc2ΞΈ1{\;} + {\;co}{{t}^2}\theta \; = \;{cs}{{c}^2}\theta
                  Here are formulas to deriving inverse trigonometric functions
                  ddx(\frac{d}{dx}(arcsin x)=11βˆ’x2 x)=\frac{1}{\sqrt{1-x^2}}
                  ddx(\frac{d}{dx}(arccos x)=βˆ’11βˆ’x2x)=\frac{-1}{\sqrt{1-x^2}}
                  ddx(\frac{d}{dx}(arctan x)=11+x2x)=\frac{1}{1+x^2}
                  ddx(\frac{d}{dx}(arccot x)=βˆ’11+x2 x)=\frac{-1}{1+x^2}
                  ddx(\frac{d}{dx}(arcsec x)=1∣x∣x2βˆ’1 x)=\frac{1}{|x| \sqrt{x^2-1}}
                  ddx(\frac{d}{dx}(arccsc x)=βˆ’1∣x∣x2βˆ’1 x)=\frac{-1}{|x| \sqrt{x^2-1}}