Home > Calculus > The special case of x⁴

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

Answer 1

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

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

Step 1: Utilize the chain rule and the product 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: compute y's derivative.

When we take the derivative of y', the second step is to once more apply the product rule and the chain rule.

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.