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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