Question

attempt 1: 10 attempts remaining. find the derivative y of the function y = 3x - x ln(x). y=

Explanation:

Step1: Apply sum - difference rule

The derivative of a sum/difference of functions is the sum/difference of their derivatives. So $y'=(3z - z\ln(z))'=(3z)'-(z\ln(z))'$.

Step2: Differentiate $3z$

Using the power - rule $(ax)' = a$ (where $a = 3$), the derivative of $3z$ with respect to $z$ is $3$.

Step3: Differentiate $z\ln(z)$

Use the product rule $(uv)'=u'v + uv'$, where $u = z$ and $v=\ln(z)$. The derivative of $u = z$ with respect to $z$ is $1$, and the derivative of $v=\ln(z)$ with respect to $z$ is $\frac{1}{z}$. So $(z\ln(z))'=1\times\ln(z)+z\times\frac{1}{z}=\ln(z) + 1$.

Step4: Calculate $y'$

$y'=(3z)'-(z\ln(z))'=3-(\ln(z)+1)=2-\ln(z)$.

Answer:

$2-\ln(z)$