How do you do the derivative of an inverse trig function?
arcsin (sqrt2/2)??
arcsin(sqrt2/2 ) is a constant. The derivative of a constant is zero.
oops mixed up 2 questionss haha
to find an arcsin you would just have to memorize the unit circle right? so arccos of -sqrt3/2 is 5pi/6?
yes, that is the easy way.
However, to find the derivative of an inverse trig function, go through the back door, as it were
y = arcsin(x)
sin y = x
cos y y' = 1
y' = 1/cos x = 1/√(1-sin2x) = 1/√(1-x2)
To find the derivative of an inverse trigonometric function, you can use the chain rule.
Let's start with the example of finding the derivative of the inverse sine function, or arcsin(x).
1. Begin with the formula of the derivative of the inverse function:
dy/dx = 1 / (dx/dy)
2. Identify the inverse sine function:
y = arcsin(x)
3. Rewrite the equation in terms of x:
x = sin(y)
4. Calculate the derivative of both sides of the equation with respect to x:
d/dx(x) = d/dx(sin(y))
5. Simplify the left side of the equation (derivative of x with respect to x is 1):
1 = d/dx(sin(y))
6. Use the chain rule, which states that d/dx(f(g(x))) = f'(g(x)) * g'(x):
d/dx(sin(y)) = cos(y) * dy/dx
7. Substitute the value of y:
1 = cos(y) * dy/dx
8. Rearrange the equation to solve for dy/dx:
dy/dx = 1 / cos(y)
9. Use the Pythagorean identity sin^2(y) + cos^2(y) = 1 to find the value of cos(y):
sin^2(y) + cos^2(y) = 1
cos^2(y) = 1 - sin^2(y)
cos(y) = sqrt(1 - sin^2(y))
10. Plug in the given value of sin(y) = sqrt(2) / 2:
cos(y) = sqrt(1 - (sqrt(2) / 2)^2)
= sqrt(1 - 1/2)
= sqrt(1/2)
= 1/sqrt(2)
= sqrt(2) / 2
11. Substitute the value of cos(y) back into the equation:
dy/dx = 1 / (sqrt(2)/2)
= 2 / sqrt(2)
= sqrt(2)
Therefore, dy/dx for arcsin(sqrt(2)/2) is sqrt(2).
You can repeat a similar process for other inverse trigonometric functions, such as arccos(x) and arctan(x), by using their respective differentiation rules and identities.