# Thread: Basic question - derivatives

1. ## Basic question - derivatives

Hi,

I'm really new to this, just trying to get my head around things.

I've been given the question: 1/sqrt(x) + 3x - x (I've attached a clearer jpeg version)

I'm using the formula: y' = b * ax ^b-1

I am getting the answer '3', through the following working out, however this does not correspond to the answer given by online derivative calculators.

--------------------------------

#1:

1/sqrt(x) + 3x - x

#2:

1/sqrt(x^1) + 3x^1 - x^1

#3:

1/sqrt(1x^1-1) + 1(3x^1-1) - 1x^1-1

#4:

1/sqrt(1x^0) + 1(3x^0) - 1x^0

#5:

1/sqrt(1) + 3 - 1

#6:

1 + 3 -1 = Ans 3

------------------------------------

Can anyone help me with this? Just need to know where or if I'm going wrong.

Thanks.
Daniel.

2. Originally Posted by ashleysmithd
Hi,

I'm really new to this, just trying to get my head around things.

I've been given the question: 1/sqrt(x) + 3x - x (I've attached a clearer jpeg version)

I'm using the formula: y' = b * ax ^b-1

I am getting the answer '3', through the following working out, however this does not correspond to the answer given by online derivative calculators.

--------------------------------

#1:

1/sqrt(x) + 3x - x

#2:

1/sqrt(x^1) + 3x^1 - x^1

#3:

1/sqrt(1x^1-1) + 1(3x^1-1) - 1x^1-1

#4:

1/sqrt(1x^0) + 1(3x^0) - 1x^0

#5:

1/sqrt(1) + 3 - 1

#6:

1 + 3 -1 = Ans 3

------------------------------------

Can anyone help me with this? Just need to know where or if I'm going wrong.

Thanks.
Daniel.

$\displaystyle \frac{1}{\sqrt{x}} + 3x - x$

$\displaystyle \frac{1}{\sqrt{x}} + 2x$

$\displaystyle x^{-\frac{1}{2}} + 2x$

$\displaystyle \frac{d}{dx}\left(x^{-\frac{1}{2}} + 2x\right)$

$\displaystyle -\frac{1}{2} x^{-\frac{3}{2}} + 2$

$\displaystyle -\frac{1}{2x\sqrt{x}} + 2$

... in future, please post calculus questions in the calculus section.