Question

Find y' and y''? `y = x^2ln(2x)`

Answer 1

`y' =2xln(2x)+x`

`y''= 2ln(2x)+3`

Explanation:

Step 1: Apply the product rule and the chain rule

Since the original equation is `x^2` times ln(2x), we can apply the product rule first. y' = `2xln(2x)+x^2*(1/(2x))*2`
(*note that the derivative of ln(x) is just `1/x` so we only need to apply the chain rule for taking the derivative of ln(2x) . )

Now we get `y' = 2xln(2x)+x` after simplifying

Step 2: take the derivative of y'

The second step is to apply the product rule and the chain rule again when we take the derivative of y'.

So y'' = `2*ln(2x) + 2x * 1/(2x) *2 + 1`

Now we get y'' = `2*ln(2x)` + 3 after simplifying

Answers:

`y' =2xln(2x)+x`

`y''= 2ln(2x)+3`