#y=tan^2(x)# Find #dy/dx#The answer is #2tan(x)sec^2(x)#. It was solved by substituting #u=tan(x)#. Now why did we not substitute #u=tan(x) for the second part? See below for more details

This is how we solved it :

#dy/dx=d/dx(u^2)d/dxcolor(red)((tan(x)))#

#=2usec^2(x)# Now substituting back #u=tan(x)#,
#2tan(x)sec^2(x)#.

My question is why didn't we substitute #tan(x)# for #u#? (The red part)

If we did it will look like this

#dy/dx=d/dx(u^2)d/dx(u)#
#=(2u)(1)#
#=2u#
#=2tan(x)#, but that is definitely not the answer. So how do I know when to substitute and when not to?

2 Answers
Jim G.
Aug 13, 2017

#"see explanation"#

Explanation:

#"note that "d/dx(u)-=(du)/dx#

#"since u is a function of x it must be differentiated with "#
#"respect to x"#

#•color(white)(x)rArrdy/dx=dy/(du)xx(du)/dxlarrcolor(blue)" chain rule"#

#"let "u=tanxrArr(du)/dx=sec^2x#

#rArry=u^2rArrdy/(du)=2u#

#rArrdy/dx=2usec^2x=2tanxsec^2x#

mason m
Aug 13, 2017

If you already said that #u=tan(x)#, then #d/dx(u)# and #d/dxtan(x)# are the same thing.

Your issue is that you said that #d/dx(u)=1#, which is not true.

#d/(du)(u)=1#, but since #d/(dcolor(red)x)(u)# differentiates with respect to #color(red)x#, not #u#, the derivative isn't #1#.

In terms of your problem, it's more helpful to write #d/dxtan(x)# because it's clear that the derivative is #sec^2(x)#.

If you write #d/dx(u)#, well, all this says is "the derivative of the function #u# with respect to #x#." Without stating the function #u#, this is as simple as it gets—#u# could be as simple as #2x+3# or as complex as #sqrtln(cos(3pi//x))#.

Related questions
Impact of this question
668 views around the world
You can reuse this answer
Creative Commons License