How do you differentiate #y = log_2 (x^4sinx)#?

Question

1749 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-28T18:23:21

# y' = (4 sin x + x cos x)/(ln 2 * xsinx)#

#y = log_2 (x^4sinx)#

first switch into natural logs as these are calculus friendly

#y = (ln (x^4sinx))/ln 2 = implies ln 2 \ y = ln (x^4sinx)#

so therefore by the chain rule

#ln2 \ y' = 1/(x^4sinx)*(x^4sinx)'#

#(x^4sinx)' = (x^4)' sin x + x^4 (sin x)'#
#= 4x^3 sin x + x^4 cos x#
#= x^3(4 sin x + x cos x)#

so the derivative is
#ln2 \ y' = 1/(x^4sinx)*x^3(4 sin x + x cos x)#
#= (4 sin x + x cos x)/(xsinx)#

so

# y' = (4 sin x + x cos x)/(ln 2 * xsinx)#