# Math Help - Derivative Question?

1. ## Derivative Question?

Hi All!

I am new to this topic and not sure if it is correct place to post this.

I know simple derivative like $x^3-x-1$. For $x^3$ we multiply it by power and subtract 1 from power which is equal to $3x^2$, x will become 1 and derivative of 1 is 0 so answer is

$3x^2-1$

Don't know how to calculate derivative of

2. Originally Posted by mrsenim
Hi All!

I am new to this topic and not sure if it is correct place to post this.

I know simple derivative like $x^3-x-1$. For $x^3$ we multiply it by power and subtract 1 from power which is equal to $3x^2$, x will become 1 and derivative of 1 is 0 so answer is

$3x^2-1$

Don't know how to calculate derivative of
$y = \frac{1}{\sqrt{x+1}}$

Chain rule works best here IMO.

Let $u = x+1$. This should be pretty easy to solve for $\frac{du}{dx}$

Thus we get $y = \frac{1}{\sqrt{u}}$.

Remembering the laws of exponents this is equal to $y = u^{-1/2}$ which differentiates in the same way as the positive integers so you can find $\frac{dy}{du}$.

Here's where the chain rule comes in:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

You should have and expression for both those terms

3. $\frac{1}{\sqrt{x + 1}} = (x + 1)^{-\frac{1}{2}}$.

Now you need to use the Chain Rule.

4. Originally Posted by Prove It
$\frac{1}{\sqrt{x + 1}} = (x + 1)^{-\frac{1}{2}}$.

Now you need to use the Chain Rule.
I tried to solve it as follows

$\frac{1}{\sqrt{x + 1}} = (x + 1)^{-\frac{1}{2}}$

Now multiplying by -1/2 and subtracting 1 from power

$= {-\frac{1}{2}}[(x + 1)^{-\frac{3}{2}}]$

$= {-\frac{1}{2}}[\frac{1}{(x + 1)^{\frac{3}{2}}}]$

Is it correct?

Don't know what is chain rule.

5. Originally Posted by mrsenim
I tried to solve it as follows

$\frac{1}{\sqrt{x + 1}} = (x + 1)^{-\frac{1}{2}}$

Now multiplying by -1/2 and subtracting 1 from power

$= {-\frac{1}{2}}[(x + 1)^{-\frac{3}{2}}]$

$= {-\frac{1}{2}}[\frac{1}{(x + 1)^{\frac{3}{2}}}]$

Is it correct?