What is the derivative of #f(x)=x/(x^2sin^2x+x^2cos^2x)#?

1 Answer
Oct 29, 2015

This is a bit of a trick or trap question.

Explanation:

You certainly could use the quotient, product and chain rules for this function, but you'll probably prefer the method below:

#f(x)=x/(x^2sin^2x+x^2cos^2x)#

# = x/(x^2(sin^2x+cos^2))# #" "# factor to leave #sin^2x+cos^2#)

# = x/(x^2(1))# #" "# #" "# #" "# (use the well known trig identity)

# = 1/x = x^-1#

#f(x) = x^-1#, so

#f'(x) = -1/x^2#.

The purpose of such a question is partly to review algebra and trigonometry, and (perhaps more importantly) to teach students that we can control how things are written. If some simplification before differentiating makes it easier, then do the simplification.

Important note sometimes a practice problem or an examination will specify a particular technique. The purpose of this is to teach and test other techniques.