Question

hw_2.1_basic derivative rules
due sunday by 11:59pm points 100.05
hw_2.1_basic derivative rules

1. submit answer practice similar

attempt 1: 10 attempts remaining.
differentiate $g(u)=sqrt{7}u+sqrt{6u}$.
answer: $g(u)=$

Explanation:

Step1: Recall derivative rules

The derivative of a sum of functions is the sum of their derivatives, i.e., $(f + g)'=f'+g'$. Also, the derivative of $au$ with respect to $u$ (where $a$ is a constant) is $a$, and the derivative of $\sqrt{au}=a^{\frac{1}{2}}u^{\frac{1}{2}}$ with respect to $u$ is $\frac{a^{\frac{1}{2}}}{2}u^{-\frac{1}{2}}$ by the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$.

Step2: Differentiate each term

For the first term $\sqrt{7}u$, since $\sqrt{7}$ is a constant, its derivative with respect to $u$ is $\sqrt{7}$. For the second term $\sqrt{6u}=\sqrt{6}u^{\frac{1}{2}}$. Using the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$, we have $\frac{d}{du}(\sqrt{6}u^{\frac{1}{2}})=\sqrt{6}\times\frac{1}{2}u^{\frac{1}{2}-1}=\frac{\sqrt{6}}{2\sqrt{u}}$.

Step3: Combine the derivatives

$g'(u)=\sqrt{7}+\frac{\sqrt{6}}{2\sqrt{u}}$

Answer:

$g'(u)=\sqrt{7}+\frac{\sqrt{6}}{2\sqrt{u}}$