How do you find the derivative of #y = arcsin(5x)#?

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Answer 1 · 2016-11-15T17:22:12

# dy/dx = 5/sqrt(1-25x^2) #

# y = arcsin(5x) #

# siny = 5x # ..... [1]

We can now differentiate implicitly to get:

# cos(y)dy/dx = 5 # ..... [2]

Using the fundamental trig identity #sin^2A+cos^2A-=1# we can write:

# sin^2(y+cos^2(y)=1#
# :. (5x)^2+cos^2(y)=1# (from [1])
# :. cos^2(y)=1-25x^2#
# :. cos(y)=sqrt(1-25x^2)#

Substituting into [2] we get:

# sqrt(1-25x^2)dy/dx = 5 #

# :. dy/dx = 5/sqrt(1-25x^2) #