Derivatives of Inverse Trig Functions

Derivatives of inverse trig functions

In the real world, trigonometric functions are used to model many trpes of motion. The derivative of these functions can be used to determine how quickly an object is accelerating, or how fast it is moving at any given time. Knowing the derivative and inverse relations of trig functions has practical applications in physics and engineering.
We will develop the basic formulas for inverse trigonometric functions. To begin, let us consider the arc sine function. We cannot find its derivative directly, but we can rewrite this in terms of sine.

This gives us x and y in terms of x and y, which is what we want since sine and inverse sine are inverse functions. X when multiplied by its original value will give us the full value back. We can apply the derivative of sine to both sides, yielding cosine
y. Next, we can use implicit differentiation to find dy dx. After that, we have 1 over the derivative of x, which is equal to 1 over cosine y.

However, we need to look at the given information in terms of x. While it may be possible to find a solution by considering x as y/1, this
approach does not yield meaningful results. The sine of y is x. If we think of x as the hypotenuse and y as the opposite side, we get that the opposite side is x. We need to find the length of this side, which is x. By the Pythagorean theorem, the cosine of an angle is equal to the adjacent side squared minus the hypotenuse squared.

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Are sine of x is adjacent over hypotenuse. This can be proven by applying the formula for arc sine: sqrt 1 – x2. The derivative of this function is 1 sqrt 1 be proven using calculus. Therefore, the derivative of arc sine is 1 over – x2 and can
the square root of 1 minus x squared.
This is the derivative of arc sine of x.
y=(sin^-1 x)
d/dx[siny=x/1]
cos y dy/dx=1
dy/dx=1/cosy
dy/dx=1/sqrt(1-x^2) =d/dx sin^-1 x/1
cosy=sqrt(1-x^2)
We can continue our work and say, "Hey, you know what? We want now the arc cosine." Based on what you know about sine and cosine, what can you do to get the arc cosine? You might have a hunch about how it works.
Let's see if your guess is right. Apply cosine to both sides. I'll go through this faster than before. The derivative of y with respect to x is equal to 1. Now you can guess better. Derivative again, derivative again.
The sine of an angle is equal to the ratio of the opposite side over the hypotenuse. This formula can be generalized to other trigonometric functions. By applying triangle trigonometry, we can find that the cosine of y is equal to x. This means that the side opposite y 1s x and the hypotenuse is 1 -x squared surprise
So the derivative of arc cosine y is -negative 1 over square 1 minus × squared. -well, this could have been a
y=cos^-1 x
d/dx cosy=x
-siny dy/dx=1
dy/dx=-1/siny [Graph] siny=sqrt(1-x^2)
d/dx cos^-1 y =-1/(sqrt(1-x^2))
We'll continue by taking the derivative of tan x. We'll use the chain rule to differentiate secant squared dx. And then we'll solve for dy/dx, obtaining dy dx = secant squared of tan x.
The derivative of the tangent function is equal to 1 divided by the square root of one plus × squared. This can be proven using basic trigonometry and the relationships between derivatives and integrals.
Derivatives of trigonometric functions, like secant squared, can be found using the identities: cosine squared = 1/2 tan squared and 1/tan squared = secant squared over cosine squared. This can also be applied to tangent functions; the derivative of arc tan x is 1 over (1 + × squared).
y=trn^-1 x
d/dx trn y= x
Sec^2 y dy/dx=1
dy/dx=1/(Sec^2 y)=cos^2 y [Graph]
dy/dx trn^-1 x =1/(1+x^2) cosy=1/(sqrt(1+x^2))
That concludes our discussion of inverce trigonometric functions and oscillations.