# Thread: [SOLVED] Differentiate √(x² + 1) with respect to x.

1. ## [SOLVED] Differentiate √(x² + 1) with respect to x.

Question:
Differentiate $\displaystyle \sqrt{(x^2 + 1)}$ with respect to $\displaystyle x$.

Attempt:

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

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

$\displaystyle \frac{dy}{du} = x^2$

$\displaystyle \frac{du}{dx} = \frac{1}{2}u^{-\frac{1}{2}}$

$\displaystyle \frac{dy}{dx} = x^2 \times \frac{1}{2}u^{-\frac{1}{2}}$

$\displaystyle \frac{dy}{dx} = x^2 \times \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}$

Are my steps right? and can you please complete the steps? thnx!

dy/dy = 2x(1/2)(x^2+1)^(-1/2)

dy/dy = 2x(1/2)(x^2+1)^(-1/2)
Why is it $\displaystyle 2x$ instead of $\displaystyle x^2$ ?

4. Hello,

Because the derivative for x²+1 is 2x

5. take for example X^n

differentiating it w.r.t x will give:
n(d/dx X)(X)^(n-1)

take for example (X+1)^n

differentiating it w.r.t x will give:
n(d/dx (X+1))(X+1)^(n-1)

take for example ((X^m)+1)^n

differentiating it w.r.t x will give:
n(d/dx ((X^m)+1))((X^m)+1)^(n-1)

in youe question n = 1/2, m=2.

6. Originally Posted by looi76
Question:
Differentiate $\displaystyle \sqrt{(x^2 + 1)}$ with respect to $\displaystyle x$.

Attempt:

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

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

$\displaystyle \frac{dy}{du} = x^2$

$\displaystyle \frac{du}{dx} = \frac{1}{2}u^{-\frac{1}{2}}$

$\displaystyle \frac{dy}{dx} = x^2 \times \frac{1}{2}u^{-\frac{1}{2}}$

$\displaystyle \frac{dy}{dx} = x^2 \times \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}$

Are my steps right? and can you please complete the steps? thnx!
What a strange notation you have there...
I would write immediately:
$\displaystyle \left(\sqrt{(x^2 + 1)}\right)' = \left((x^2 + 1)^{\frac{1}{2}}\right)'=\frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}\cdot{2x}=x (x^2 + 1)^{-\frac{1}{2}}=\frac{x}{\sqrt{x^2 + 1}}$