Please help me with this math question.?

Question

Find the derivative of y with respect to x.

y= x ln (x) / 4 + ln (x)

Answer 1

dy/dx = u*dv/dx + v*du/dx

SO- 0.25x ln(x) becomes- 0.25*ln(x) + 0.25 x/x

The whole thing becomes-
dy/dx = ln(x)/4 + 1/x + 1/4

Answer 2

*sigh* You guys need to start using parentheses. Is the final ln x in the denominator or is it a separate term? I'm going to assume it's in the denominator.

d/dx (x ln(x)/(4 + ln(x))

First off, you'll need the quotient rule

= [(4 + ln(x)) d/dx(x ln(x)) - x ln(x) d/dx(4 + ln(x)]/(4 + ln(x))^2

Now the product rule for the first term in the denominator:

d/dx (x ln(x)) = x(1/x) + ln(x) = 1 + ln(x)

So df/dx = [(4 + ln(x))(1 + ln(x)) - x ln(x)(1/x)]/(4 + ln(x))^2

Obviously, the x and 1/x in the second term in the numerator cancel, so we're at

df/dx = [(4 + ln(x))(1 + ln(x)) - ln(x)]/(4 + ln(x))^2

If you multiply out the first term, it may simplify, but since the rest of the problem is just algebra, I'll leave you to it.

Answer 3

i got
((1+ln(x))/4)+(1/x)

so that's (1+ln(x)) all over four, then add 1/x to that.

Answer 4

y'=(ln(x))^2+(4ln(x)+4)/(ln(x)+4)^2

I'm sure this is right..

Answer 5

i don't know