How do you differentiate #f(x)=sinx^2/x^2#?

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Answer 1 · 2017-05-22T18:06:52

For #f(x) = u/v#, we have #f'(x) =(u'v-uv')/v^2#

In this quotient, #u = sinx^2#, which I will take to be #sin(x^2)# (rather than #(sinx)^2# which is a different function).

This makes #u' = cos(x^2) * 2x = 2xcos(x^2)# #" "# (chain rule)

and #v = x^2#, so #v' = 2x#.

#f'(x) =(u'v-uv')/v^2#

# = ((2xcos(x^2))(x^2)-((sin(x^2)(2x))))/(x^2)^2#

# = (2x^3cos(x^2) - 2xsin(x^2))/x^4#

Every term has a factor of #x#, so we can simplify the answer.

# = (cancel(x)(2x^2cos(x^2) - 2sin(x^2)))/(cancel(x) x^3)#

# = (2x^2cos(x^2) - 2sin(x^2))/x^3#