Question

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

Answer 1

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

Explanation:

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))`