Finding the derivatives of the inverse trigonometric functions involves using _____ and the derivatives of regular trigonometric functions.

Question

Choices:

A. Chain rule

B. explicit differentiation

C. Implicit differentiation

D. All of these

Answer 1

to differentiate inverse trig. functions

all three of listed techniques are used.

let y=sin(sin^-1(x)

that is same as y=x

implicit on left hand side, explicit on right hand side, have y'=1

y'=dy/dx=[d(sin(sin^-1(x))/d(sin^-1(x))]*[d(sin^-1(x))/dx),,,,,here is the chain rule

Note the first term is cos(sin^-1(x)) and the second term is the derivative of sin^-1(x) with respect to x

Answer 2

y = arcsin(x)

means siny = x

The derivative of siny = x is cosy * dy/dx = 1

dy/dx = 1/cosy

That's implicit differentiation.

Look at this image of the triangle: https://imgur.com/BUDR8Cx

If siny = x, then y is one of the acute angles, the side opposite y is x, and the hypotenuse is 1. siny = opp/hyp = x/1.

Pythagorean Theorem finds the other side, the adjacent side to angle y.

a^2 + x^2 = 1^2

a^2 = 1-x^2

a = sqrt(1-x^2)

cosy = adh/hyp = sqrt(1-x^2)/1 = sqrt(1-x^2)

dy/dx = 1/cosy = 1/sqrt(1-x^2)

C