How do you differentiate #log_2(x^2sinx) #?

Question

How do you differentiate #log_2(x^2sinx)   #?

Charles Arocha · Accepted Answer

#2/(ln(2) x ) + cos(x) / (ln(2) sin(x))#

First, I would strongly suggest to use the logarithmic rules to simplify your expression.

Remember the logarithmic rules:

#log_a (x * y ) = log_a(x) + log_a(y)# and #log_a(x^n) = n * log_a(x) #

With these rules, you can simplify as follows:

#f(x) = log_2(x^2 sin x ) = log_2(x^2) + log_2(sin x)# #color(white)(xxx)= 2 * log_2(x) + log_2(sin x)#

Now, we know the derivative of #ln(x)# which is #1/x#. However, you don't have #ln(x)# but #log_2(x)# here, so it would be good to convert the logarithmic expressions to #ln(x)#.

To transform the base of an logarithm, you can use the following rule: # log_b(x) = log_a(x) / log_a(b)#

In our case, it means that

#log_2(x) = ln(x) / ln(2)#

holds. Thus, we can transform further:

#f(x) = 2/ln(2) ln(x) + ln(sin(x)) / ln(2) = 2/ln(2) ln(x) + 1/ ln(2) * ln(sin(x)) #

Now, it's easy to differentiate the first part, and for the second part you just need the chain rule:

#f'(x) = 2/ln(2) * 1/x + 1/ln(2) * 1/sin(x) * cos(x) #

#color(white)(xxxx) = 2/(ln(2) x ) + cos(x) / (ln(2) sin(x))#

Emily Ammerman · Answer

undefined To differentiate the function log base 2 of (x^2 * sin(x)), you would apply the chain rule and the product rule. The derivative is:

[2 * x * sin(x) / (x^2 * sin(x) * ln(2))] + [2 * ln(2) / (x^2 * sin(x))]