Question

question
find $\frac{d}{dx}(-3x - 3x^{1/6}+1)$.
provide your answer below:
$\frac{d}{dx}(-3x - 3x^{1/6}+1)=square$

Explanation:

Step1: Apply sum - difference rule of derivatives

The derivative of a sum/difference of functions is the sum/difference of their derivatives, i.e., $\frac{d}{dx}(u - v+w)=\frac{d u}{dx}-\frac{d v}{dx}+\frac{d w}{dx}$. Here $u = - 3x$, $v = 3x^{1/6}$, $w = 1$. So $\frac{d}{dx}(-3x - 3x^{1/6}+1)=\frac{d}{dx}(-3x)-\frac{d}{dx}(3x^{1/6})+\frac{d}{dx}(1)$.

Step2: Apply constant - multiple rule of derivatives

The constant - multiple rule states that $\frac{d}{dx}(cf(x))=c\frac{d}{dx}(f(x))$, where $c$ is a constant. So $\frac{d}{dx}(-3x)-\frac{d}{dx}(3x^{1/6})+\frac{d}{dx}(1)=-3\frac{d}{dx}(x)-3\frac{d}{dx}(x^{1/6})+\frac{d}{dx}(1)$.

Step3: Apply power rule of derivatives

The power rule is $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $y = x$, $n = 1$, so $\frac{d}{dx}(x)=1$. For $y=x^{1/6}$, $n=\frac{1}{6}$, so $\frac{d}{dx}(x^{1/6})=\frac{1}{6}x^{\frac{1}{6}-1}=\frac{1}{6}x^{-\frac{5}{6}}$. And $\frac{d}{dx}(1)=0$ (since the derivative of a constant is 0). Then $-3\frac{d}{dx}(x)-3\frac{d}{dx}(x^{1/6})+\frac{d}{dx}(1)=-3\times1-3\times\frac{1}{6}x^{-\frac{5}{6}}+0$.

Step4: Simplify the expression

$-3\times1-3\times\frac{1}{6}x^{-\frac{5}{6}}+0=-3-\frac{1}{2}x^{-\frac{5}{6}}$.

Answer:

$-3-\frac{1}{2}x^{-\frac{5}{6}}$