Deriatives

[dave turbo]

Taking the deriative of this function
Is this correct?

y = 4x^3 + x^2/2 + 6

y' = (3)4x^2 + (2)2x + 0

y' = 12x^2 + 4

 

[math-is-fun]

Taking the deriative of this function
Is this correct?

y = 4x^3 + x^2/2 + 6

y' = (3)4x^2 + (2)2x + 0

y' = 12x^2 + 4

x^2/2 = x, as x^2/2 is equivalent to squaring x and raising x to the power of 2; and of the course, the derivative of x is 1 (so y' would = 12x^2 + 1). but, just in case I'm reading your post wrong, (x^2)/2 derives into 2x/2, which reduces to x (so y' would = 12x^2 + x). I'm learning calc. I myself right now, but I'm pretty confident in my answer. Please correct me if I'm wrong though and best of luck!

-Matthew

 

[Deleted member 4993]

x^2/2 = x, as x^2/2 is equivalent to squaring x and raising x to the power of 2; and of the course, the derivative of x is 1 (so y' would = 12x^2 + 1). but, just in case I'm reading your post wrong, (x^2)/2 derives into 2x/2, which reduces to x (so y' would = 12x^2 + x). I'm learning calc. I myself right now, but I'm pretty confident in my answer. Please correct me if I'm wrong though and best of luck!

-Matthew

x^2/2 is equivalent to $\displaystyle \dfrac{x^2}{2}$ following PEMDAS (or PEDMAS)

Otherwise your post is correct

 

[Deleted member 4993]

Taking the deriative of this function
Is this correct? .................No

y = 4x^3 + x^2/2 + 6

y' = (3)4x^2 + (2)2x + 0

y' = 12x^2 + 4

As Mathew has shown above:

y = 4x^3 + x^2/2 + 6

y' = (3)4x^2 + (1/2)2x + 0

y' = 12x^2 + x

 

[Quaid]

x^2/2 = x

Oops. That's not what you want to say.

The equation above is only true when x = 0 or x = 2.

Here is what I think you were trying to say.

d/dx [x^2/2] = x

x^2/2 is equivalent to squaring x and raising x to the power of 2

In this statement, replace the word and with the word or. Also, you forgot to mention the division by 2. Without these corrections, the word 'equivalent' is not correct.

Cheers