If #sin^-1(x) +sin^-1(y)=pi/2#, then what will #dy/dx# be?

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Answer 1 · 2018-01-30T18:07:44

Let's have a look.

This problem involves both differentiation as well as inverse trigonometric functions.

Now given that,

#sin^(-1)x+sin^(-1)y=pi/2#

#:.sin^(-1)x=pi/2-sin^(-1)y#

#:.sin^(-1)x=cos^(-1)y#

#:.y=cossin^(-1)x#

#:.dy/dx=-sinsin^(-1)x.d/dx(sin^(-1)x)#

#:.dy/dx=-x1/sqrt(1-x^2)#

#:.color(red)(dy/dx=-x/sqrt(1-x^2))#.

hope it Helps :)

Answer 2 · 2018-01-30T19:25:23

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

We have:

# sin^(-1)x + sin^(-1)y = pi/2 #

Using the standard result:

# d/dx sin^(-1) x = 1/sqrt(1-x^2) #

we can implicitly differentiate the first equation, giving:

# 1/sqrt(1-x^2) + 1/sqrt(1-y^2) dy/dx = 0 #

# :. 1/sqrt(1-y^2) dy/dx = - 1/sqrt(1-x^2) #

# :. dy/dx = - sqrt(1-y^2)/sqrt(1-x^2) #