What is the derivative of #y = log_2 (x^4sinx)#?

1 Answer
Aug 24, 2015

#d/dx(log_2(x^4sinx))(4 +xcotx)/(xln2)#

Explanation:

I like to find the general form of an equation before taking the derivative, so I can see what rules I'm going to have to use. In this case we have;

#y=log_2(x^4sinx) = f(g(x)h(x))#

#f=log_2#
#g(x)=x^4#
#h(x)=sinx#

Since #f# is a function of #g# and #h#, we are going to need the chain rule. #f# is logarithmic, so it follows the form;

#d/(dx)log_a(x) = d/dx ln(x)/ln(a) = 1/(xlna)#

Applying the chain rule we get;

#d/dx log_2(g(x)h(x)) = 1/(g(x)h(x)ln2) d/dx(g(x)h(x))#

Now we need to apply the product rule to solve the last part. The product rule tells us;

#d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x)#

So our general solution is;

#1/(g(x)h(x)ln2)(f'(x)g(x) + f(x)g'(x))#

#=(g'(x)h(x) + g(x)h'(x))/(g(x)h(x)ln2)#

Now we need to find #g'# and #h'#.

#d/dx x^4 = 4x^3#

#d/dx sinx = cosx#

The rest is plugging in and simplifying.

#(4cancel(x^3)cancel(sinx) + x^cancel(4)cancel(cosx)^cotx)/(x^cancel(4)cancel(sinx)ln2) = (4 +xcotx)/(xln2)#