# Question #4686f

Mar 3, 2017

${h}^{'} \left(x\right) = \frac{6 a}{c}$

#### Explanation:

Simply put, a constant is a number that does not vary. A constant could be equal to $0$, but it could be equal to $1$, $10$, $\sqrt{2}$, $\pi$, and so on.

In this case, $a$, $b$, and $c$ are defined as constants, which implies that their value does not vary. In other words, it doesn't matter what happens to the function, $a$, $b$, and $c$ will always have the same value.

In this particular instance, it's not important what that value will be, what matters is that it does not change.

The idea with constants is that their derivative is always equal to $0$.

$\frac{d}{\mathrm{dx}} \left(\text{constant}\right) = 0$

Now, your function looks like this

$h \left(x\right) = \frac{6 a x + 10 b}{c}$

You can rearrange this function as

$h \left(x\right) = \frac{6 a x}{c} + \frac{10 b}{c}$

Since the term $\frac{10 b}{c}$ consists entirely of constants, it will also be a constant. The same can be said about $\frac{6 a}{c}$.

So your function can be written as

$h \left(x\right) = {\text{constant"_1 * x + "constant}}_{2}$

Here

$\text{constant"_1 = (6a)/c" }$ and ${\text{ " "constant}}_{2} = \frac{10 b}{c}$

According to the sum rule, which tells you that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left[f \left(x\right) + g \left(x\right)\right] = \frac{d}{\mathrm{dx}} \left[f \left(x\right)\right] + \frac{d}{\mathrm{dx}} \left[g \left(x\right)\right]}}}$

you can differentiate this function by going

${\overbrace{\frac{d}{\mathrm{dx}} \left[h \left(x\right)\right]}}^{\textcolor{b l u e}{= {h}^{'} \left(x\right)}} = \frac{d}{\mathrm{dx}} \left(\frac{6 a}{c} \cdot x\right) + \frac{d}{\mathrm{dx}} \left(\frac{10 b}{c}\right)$

Since the derivative of a constant is always equal to $0$, you will have

${h}^{'} \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{6 a}{c} \cdot x\right) + 0$

${h}^{'} \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{6 a}{c} \cdot x\right)$

You can separate the constant term from the variable to get

${h}^{'} \left(x\right) = \frac{6 a}{c} \cdot \frac{d}{\mathrm{dx}} \left(x\right)$

Now all you have to do is use the power rule, which tells you that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\frac{d}{\mathrm{dx}} {x}^{n} = n \cdot {x}^{n - 1}}}}$

Here $n$ is also a constant. In your case, you will have

$h ' \left(x\right) = \frac{6 a}{c} \cdot 1 \cdot {x}^{\left(1 - 1\right)}$

${h}^{'} \left(x\right) = \frac{6 a}{c} \cdot 1$

Therefore,

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{h}^{'} \left(x\right) = \frac{6 a}{c}}}}$