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

2 Answers
Dec 18, 2016

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

Explanation:

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

Dec 18, 2016

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

Explanation:

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)#