How do you find #(dy)/(dx)# given #x=sqrt(tany)#?

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Answer 1 · 2016-12-18T08:15:42

#dy/dx=2sqrttany/sec^2y#

# x=sqrttany #
# d/dx x =d/dx sqrttany #
# 1=1/(2sqrttany) sec^2y dy/dx #
# dy/dx=2sqrttany/sec^2y #

Answer 2 · 2016-12-18T08:18:39

The answer is #=(2x)/(1+x^4)#

We need

#cos^2x+sin^2x=1#

#1+tan^2x=sec^2x#

#(tanx)'=sec^2x#

Therefore,

#x=sqrttany#

Squaring both sides,

#x^2=tany#

Differentiating both sides,

#2x=sec^2ydy/dx#

#sec^2y=1+tan^2y=1+x^4#

Therefore,

#dy/dx=(2x)/(1+x^4)#