# Question #4686f

##### 1 Answer

#### Explanation:

Simply put, a **constant** is a number that *does not vary*. A constant *could* be equal to

In this case,

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

#d/dx("constant") = 0#

Now, your function looks like this

#h(x) = (6ax + 10b)/c#

You can rearrange this function as

#h(x) = (6ax)/c + (10b)/c#

Since the term *constants*, it will also be a constant. The same can be said about

So your function can be written as

#h(x) = "constant"_1 * x + "constant"_2#

Here

#"constant"_1 = (6a)/c" "# and#" " "constant"_2 = (10b)/c#

According to the **sum rule**, which tells you that

#color(blue)(ul(color(black)(d/dx[f(x) + g(x)] = d/dx[f(x)] + d/dx[g(x)])))#

you can differentiate this function by going

#overbrace(d/dx[h(x)])^(color(blue)(=h^'(x))) = d/dx((6a)/c * x) + d/dx((10b)/c)#

Since the derivative of a constant is always equal to

#h^'(x) = d/dx((6a)/c * x) + 0#

#h^'(x) = d/dx((6a)/c * x)#

You can separate the constant term from the variable to get

#h^'(x) = (6a)/c * d/dx(x)#

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

#color(blue)(ul(color(black)(d/dxx^n = n * x^(n-1))))#

Here

#h'(x) = (6a)/c * 1 * x^((1-1))#

#h^'(x) = (6a)/c * 1#

Therefore,

#color(darkgreen)(ul(color(black)(h^'(x) = (6a)/c)))#