If $y = 6x\ln(x)$, find the following: $y''$
Calculated $y'$ to: $$\frac{6x}{x} + 6\ln(x)$$
Isn't $$y'' = 6x(1) + \frac{x(6)}{x^2} + \frac{6}{x}$$
I don't understand why that is wrong?
$\endgroup$5 Answers
$\begingroup$You are almost there. Your $y'$ is correct. You need to simplify it further, to get: $$y'=6+6\ln x$$ and so we get $$y''=0+\frac{6}{x}=\frac{6}{x}$$
$\endgroup$ $\begingroup$You asked for what was wrong with your work, and this is an attempt to explain that.
You say: calculated $y'$ to: $$\frac{6x}{x} + 6\ln(x)$$
The derivative of $\frac{1}{x}$ is $\frac{-1}{x^2}$
Therefore, if you don't simplify $\frac{6x}{x}$ to $6$ like the others have done here, you will get:
$$y'' = 6 (1) \bigg(\frac{1}{x}\bigg) + 6 (x)\bigg(\frac{-1}{x^2}\bigg) + \frac{6}{x}$$
$$= \bigg(\frac{6}{x}\bigg) - \bigg(\frac{6}{x}\bigg) + \frac{6}{x}$$
$$y'' = \frac{6}{x}$$
$\endgroup$ 4 $\begingroup$$y' = 6(ln(x)+1))$
$y'' = \frac{6}{x}$
$\endgroup$ $\begingroup$$$ y'=(6x/x)+6\log(x)=6+6\log(x) $$
$$ y''=6/x $$
$\endgroup$ $\begingroup$Use product rule to get $\displaystyle y' = 6\ln(x)+\frac{6x}{x}=6\ln(x)+6$
Then differentiate again to get $\displaystyle \boxed{y''=\frac{6}{x}}$.
$\endgroup$
