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

1 Answer

#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#