1. $\frac{d}{dx}3x^2$

   To do this one, we use the power rule on the $x^2$ part, and get $2x$. However, we are also multiplying by $3$, so the answer is multiplied by $3$ as well. Hence the answer is $3(2x)=\boxed{{\displaystyle 6x}}$.

2. $\frac{d}{dx}{x}^3+x$

   By the power rule, we find $\frac{d}{dx}{x}^3=3x^2$, and $\frac{d}{dx}x$ is $\frac{d}{dx}{x}^1$ which becomes $1x^0$ by the power rule, which is $1$. By the addition rule, we have $\frac{d}{dx}{x}^3+x=\boxed{{\displaystyle 3x^2+1}}$.

3. $\frac{d}{dx}2x^3+5$

   You take the derivative of $x^3$ and you have $3x^2$. Times by $2$, that leaves $6x^2$. Okay, about the five? It is tempting to leave the five put, but actually $\frac{d}{dx}5=0$. Why? Well, it’s a constant, so it does not affect the slope. Hence we get $\boxed{{\displaystyle 6x^2}}$.

To apply the power rule in this case, we need to first multiply by the exponent ($3/2$), then subtract one from the exponent $3/2-1=1/2$. Then we have

$\frac{d}{dx}{x}^{3/2}=\frac{3}{2}{x}^{1/2}=\boxed{{\displaystyle \frac{3}{2}{x}^{1/2}}}$

In this problem, we just worry about the $x^{-2}$ to start with. We multiply by the exponent $(-2)$, and then subtract one from that to get $-2-1=-3$. Then we have

$\frac{d}{dx}7x^{-2}=7(-2x^{-3})=\boxed{{\displaystyle -14x^{-3}}}$

Find

1. $\frac{d}{dx}\sqrt{x}$

   This problem is much easier if we can rewrite the $\sqrt{x}$. This is the same thing as $x^{1/2}$, and hence we have

   $\frac{d}{dx}\sqrt{x}=\frac{d}{dx}{x}^{1/2}=\frac{1}{2}{x}^{1/2-1}=\boxed{{\displaystyle \frac{1}{2}{x}^{-1/2}}}.$

2. $\frac{d}{dx}5\sqrt[3]{x}$

   To solve this, it really helps to rewrite $\sqrt[3]{x}$ as $x^{1/3}$. Once you do that, this prolem is just the power rule and constant multiple rule:

   $\frac{d}{dx}5\sqrt[3]{x}=5x^{1/3}=5\left(\frac{1}{3}\right)x^{1/3-1}=\boxed{{\displaystyle \frac{5}{3}{x}^{-2/3}}}.$

3. $\frac{d}{dx}\sqrt[5]{x^3}+x^2+7$

   Focus on the easy parts first: we know $\frac{d}{dx}{x}^2=2x$, and we know that $\frac{d}{dx}7=0$. So we just need to figure out the $\frac{d}{dx}\sqrt[5]{x^3}$. What is this? Well, we can rewrite this as $\frac{d}{dx}{x}^{3/5}$. So now it is just the power rule, and we multiply by $3/5$ and subtract to get $-2/5$. Hence $\frac{d}{dx}{x}^{3/5}=\frac{3}{5}{x}^{-2/5}$. Putting it all together:

   $\frac{d}{dx}\sqrt[5]{x^3}+x^2+7=\boxed{{\displaystyle \frac{3}{5}{x}^{-2/5}+2x}}.$

1. $\frac{d}{dx}\frac{1}{x}$

   We can rewrite $\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}\frac{1}{x^1}$ as $\frac{d}{dx}{x}^{-1}$. Now we apply the power rule:

   $\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}{x}^{-1}=\boxed{{\displaystyle -1x^{-2}}}.$

2. $\frac{d}{dx}\frac{4}{x^2}$

   We can think of $4$ as just a constant out front, and hence we want $\frac{d}{dx}4\left(\frac{1}{x^2}\right)$. We can then rewrite $\frac{1}{x^2}$ as $x^{-2}$. And we want $\frac{d}{dx}4(x^{-2})$. Using the power rule, we multiply by $-2$ and subtract one, and we have

   $\frac{d}{dx}\frac{4}{x^2}=\frac{d}{dx}4x^{-2}=\boxed{{\displaystyle -8x^{-3}}}.$

3. $\frac{d}{dx}\frac{1}{\sqrt{x}}$

   This combines the fractional and denominator stuff. We first rewrite $\sqrt{x}$ as $x^{1/2}$:

   $\frac{d}{dx}\frac{1}{\sqrt{x}}=\frac{d}{dx}\frac{1}{x^{1/2}}.$

   We then rewrite as a negative fractional exponent.

   $\frac{d}{dx}\frac{1}{\sqrt{x}}=\frac{d}{dx}{x}^{-1/2}$

   Finally, we use the power rule.

   $\frac{d}{dx}\frac{1}{\sqrt{x}}=\boxed{{\displaystyle -\frac{1}{2}{x}^{-3/2}}}.$