Question

What is the derivative of `y=log_4 2x`?

Answer 1

`y' = 1/(xln4)`

Explanation:

`y = log_4 2x`

`y = log_4 2 + log_4x`

`y' = 0 + d/dx(log_4 x)`

Change of base (or, perhaps definition of `log_4x` depending on your course) gives:

`log_4 x = lnx/ln4`

`ln4` is constant so the derivative is

`d/dx(lnx/ln4) = 1/ln4 d/dx (lnx) = 1/ln4 * 1/x = 1/(xln4)`

`y' = 0+1/(xln4) = 1/(xln4)`

Answer 2

The answer is `dy/dx=1/(x*ln(4))`.

Explanation:

This solution involves a combination of using both the derivative of a logarithm and the chain rule.

Step 1. Break the problem into two parts:

Part A: `f(u)=log_4(u)` and
Part B: `u=g(x)=2x`

Step 2. Find the derivative of each part.

Part A: `f'(u) = 1/(u*ln(4))`. Recall that `u=2x`.
Part B: `(du)/(dx)=g'(x)=2`

Step 3. Chain Rule multiples the two derivatives. Remember to replace your temporary `u` with it's `g(x)` expression.

Chain Rule: `dy/dx=f'(u)*g'(x) = f'(g(x))*g'(x) = (1/(2xln(4)))*2`.

Finally, cancel the `2` from both the numerator and denominator for the final answer:

`dy/dx=(1/(cancel(2)xln(4)))*cancel(2)=1/(xln(4))`