How do you differentiate y = 5 x^3 + 1/2 x^2 + 3x -4

FIrslty to avoid confusion:

x = the variable x

X = multiplication.

When you differentiate an equation, the derivative (answer) is referred to as dy/dx.

For a polynomial as in this question you take each term individually and differentiate that.

Let's first do this for a generic term ax^b and we will then apply it to thisquestion​​:

The formula for the derivative of ax^b is bax^b-1​, which means you take the degree of x (whatever power it is being raised to, which in this case is b) and multiply it by the coefficient of x (which in this case in a) which provides the ba part of the solution. You then reduce the degree of x by one so it becomes x^b-1 instead of just x^b. Therefore when you put the two parts together you get the final derivative as bax^b-1.​

Now let's look at the first term of our equation : 5x³​.

The degree here is 3 so the first thing you do is multiply 3 by the coefficient of x which is 5 giving you 3x5 which is 15​.

Next, you reduce the degree of x by 1. It was x³ so now it is x^3-1 which is x²​.

Now you can put the two parts together to give the derivative of 5x³ being 15x²​.

If you do the same thing for the next term, then you get (2x1/2)x^2-1​ which is x.

​For the third time the derivative is quite nice, if you just have a number followed by x such as 3x as we have here, to get the derivative you just remove the x, meaning the derivative is 3.

​For the final term of the equation we just have a number. Whenever you differentiate a number on its own, the derivative is 0.

So now the final thing left to do is put our answer together. You need to take each of our 4 derivatives and simply add them together.

This would give you: 15x²​ + x + 3 + 0

Meaning the final answer dy/dx = 15x²​ + x + 3.