Question

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

Answer 1

`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)`