What is the implicit derivative of #1=xsiny#?

1 Answer
Dec 7, 2015

#dy/dx=-tan(y)/x#

Explanation:

If you assume #y# is a function of #x# in the equation #1=xsin(y)#, then you need the Product Rule and Chain Rule to differentiate the right-hand side with respect to #x# to get:

#0=d/dx(x) * sin(y) + x * d/dx(sin(y))=1 * sin(y) + x * cos(y) * dy/dx#

This can then be solved for #dy/dx# by subtracting #sin(y)# from both sides and then dividing both sides by #xcos(y)# to get:

#dy/dx=(-sin(y))/(x cos(y))=-tan(y)/x#.

A way we can check this answer is to actually solve the original equation for #y# as a function of #x#:

#1=xsin(y) Rightarrow sin(y)=1/x Rightarrow y=sin^{-1}(1/x)=arcsin(1/x)#'

Therefore,

#dy/dx=1/sqrt(1-(1/x)^2) * -x^{-2}=-1/sqrt(x^4-x^2)#

Since #tan(y)=tan(sin^{-1}(1/x))=1/sqrt{x^2-1}# (draw an appropriate right triangle to help you see this), it follows that

#-tan(y)/x=-1/(xsqrt{x^{2}-1})=-1/sqrt(x^4-x^2)#