How do you find the derivitive of Inverse trig function y = csc^-1(x^2+1)?

1 Answer
Aldrich
Sep 12, 2015

csc^-1(x^2 + 1) = (-2)/(x^2 + 1)

Explanation:

If you don't know how to get the derivative of inverse trigonometric functions, you can obtain the formula using implicit differentiation:

Let y = csc^-1 x.
x = csc y
d/dx x = d/dx csc y

1= dy/dx * -csc y cot y

(-1)/(csc y cot y)= dy/dx

Use a fundamental identity to express all trig functions as the one you are differentiating, in this case: cot^2 theta + 1 = csc^2 theta:

(-1)/(csc y sqrt(csc^2 y - 1) )= dy/dx

Use x = csc y:

(-1)/(x sqrt(x - 1) )= dy/dx

Therefore color(blue)(d/dx csc^(-1) x = (-1)/(x sqrt(x - 1) ))

To simplify d/dx csc^-1(x^2 + 1) we can use the chain rule:
(f @ g)'(x) = f'(g(x)) * g'(x)

d/dx csc^-1(x^2 + 1)
= (-1)/((x^2 + 1) sqrt((x^2 + 1)- 1) ) d/dx (x^2 + 1)
= (-1)/((x^2 + 1) sqrt(x^2)) * (2x)
= (-2x)/((x^2 + 1) x)
= (-2)/(x^2 + 1)

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