﻿ ﻿DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS - Differentiation of Transcendental Functions - The Calculus Primer

## The Calculus Primer (2011)

### Chapter 19. DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS

5—14.The Derivative of Arc Sin v and Arc Cos v. It will be recalled that the inverse trigonometric functions may be written as follows:

ifα = sin β,thenβ = arc sin α;

ifα = cos β, thenβ = arc cos α; etc.

Now let

y = arc sin v,

orv = sin y.(1)

Differentiating equation (1) with respect to y by  (§5—9): From equation (2), by using the formula for the derivative of an inverse function (§5—2), we have: But v is a function of x; hence, using the formula for the derivative of a function of a function (§5—1): Substituting in (4) from (3): From trigonometry Substituting in (5): By proceeding in the same manner, the reader may verify the formula for the derivative of arc cos v: 5—15.Derivative of Arc Tan v and Arc Cot v. The general procedure is the same as for the derivation given in §5—14.

Lety = arc tan v,

orv = tan y.

Differentiating with respect to y: By trigonometry, sec2 y = 1 + tan2 y = 1 + v2. Therefore In the same manner, the reader may verify 5—16.Derivative of Arc Sec v and Arc Csc v.

Lety = arc sec v,

orv = sec y.

Differentiating with respect to y: By trigonometry, and sec y = v, by hypothesis. In the same manner, the reader may verify EXAMPLE 1.Differentiate y = arc sin 2x. EXAMPLE 2.Differentiate y = arc tan . EXERCISE 5—6

Differentiate:

1. y = arc sin x2 4. y = x arc sin x 5—17.Summary of Formulas. For the reader’s convenience, we summarize below the formulas for differentiating transcendental functions, as was done for algebraic functions in §3—8.   5—18.Successive Differentiation. When discussing velocity and acceleration in §4—7, we learned how to find successive derivatives of algebraic functions. We shall now apply this idea to transcendental functions.

EXAMPLE 1.Find the fourth derivative of y = ex + 2x3. EXAMPLE 2.Find the third derivative of y = sin 2x. 5—19.Successive Differentiation of Implicit Functions. Suppose we wish to find the second derivative of y with respect to x in the equation of the ellipse b2x2 + a2y2 = a2b2. We proceed to differentiate with respect to x: Now differentiate again, remembering that y is a function of x: Substituting the value of from (1) in equation (2): But a2y2 + b2x2 = a2b2; therefore EXAMPLE 1.Find the second derivative with respect to x of

x2 + y = xy. Differentiating again: Substituting for from (1): EXAMPLE 2.Find the second derivative with respect to x of

ex = sin y. Differentiating again: Substituting the value of from (1): 5—20.The nth Derivative of a Product. It is sometimes useful to express the nth derivative of the product of two variables in terms of the variables and their successive derivatives. For example, if u and v are functions of x,then By differentiating again, the reader can verify that By mathematical induction, it can be shown, in general, that This is known as Leibniz’s Formula for the nth derivative of a product. It will be seen that the numerical coefficients follow the same law as those of the binomial theorem, and that the indices of the derivatives correspond to the exponents in the binomial expansion. The correspondence can be made complete as follows: just as the first derivatives and can be considered as and , so u and v (the variables themselves) may be considered as and .

EXERCISE 5—7

1. Find the second derivative of

y = ex sin x.

2. Find the third derivative of

y = log sin x.

3. Find the fourth derivative of

(a) y = x3 log x

(b) y = sin ax

4. Find for the equation x2 + y2 = k2.

5. Find for y2 = 4px.

6. Find for y2 + x = y.

7. Find, by Leibniz’s formula, the third derivative of y = exx2. (Let u = ex, v = x2.)

EXERCISE 5—8

Review

Differntiate: 2. y = et(1 + t2)

3. y = log (2 – 3x2)

4. y = cez

5. y = log (log x)

6. y = elog x2 8. y = ex log x

9. y = ecos 2x

10. y = log sin x

11. y = tan x + tan2 x 13. y = ex sin x 16. y = log cos2 x ﻿