What is the implicit derivative of #1=tanx-y^2#?

1 Answer
Nov 21, 2015

#2*y*dy/dx = 1/cos^2(x) #

Explanation:

y is an (unknown) function of x

Derive the left hand side
#d/dx(1) = 0#

Derive the right hand side

#d/dx(tan(x) - y^2) =d/dx(tan(x) - d/dx(y^2)#
Sum rule in differention

#tan(x) = sin(x)/cos(x) #
#d/dx(tan(x)) = d/dx(sin(x)/cos(x)) #
#d/dx(tan(x)) =(sin(x)*sin(x) - (-cos(x))*cos(x))/cos(x)^2 #
#d/dx(tan(x)) =(sin^2(x) +cos^2(x))/cos^2(x)#
#sin^2(x) +cos^2(x) = 1#
#d/dx(tan(x)) = 1/cos^2(x)#

#d/dx(y^2) = 2*y*d/dx(y) = 2*y*dy/dx #
The chain rule.

d/dx(tan(x)) = 1/cos^2(x) - 2ydy/dx #

Left hand side = right hand side

#0 = 1/cos^2(x) - 2*y*dy/dx#

Rewrite

#2*y*dy/dx = 1/cos^2(x) #