# Math Help - Differentiation

1. ## Differentiation

I am unsure if I have done the following correctly

The question: Differentiate $y=\sqrt{x^2 + \cosh^4x}$

I get my answer as $\frac{dy}{dx} = \frac{1}{2(x^2 + \cosh^4)^2}$

Thanks for any help or suggestions, they are greatly appreciated.

Beard

2. Here, the Chain Rule may be used along with the Power Rule:

$\frac{dy}{dx}=\frac{d}{dx}\sqrt{x^2+\cosh^4 x}=\frac{1}{2\sqrt{x^2+\cosh^4 x}}\frac{d}{dx}(x^2+\cosh^4 x).$

3. Originally Posted by Scott H
Here, the Chain Rule may be used along with the Power Rule:

$\frac{dy}{dx}=\frac{d}{dx}\sqrt{x^2+\cosh^4 x}=\frac{1}{2\sqrt{x^2+\cosh^4 x}}\frac{d}{dx}(x^2+\cosh^4 x).$
So is this saying that I was right?

4. Originally Posted by Scott H
Here, the Chain Rule may be used along with the Power Rule:

$\frac{dy}{dx}=\frac{d}{dx}\sqrt{x^2+\cosh^4 x}=\frac{1}{2\sqrt{x^2+\cosh^4 x}}\frac{d}{dx}(x^2+\cosh^4 x).$
I have tried redo this to see if i could simplify it in a better way but I am unsure about the answer I got

$\frac{1}{8(x^2 + 2\sinh^{3}(x) + 2\sinh^{6}h(x))}$

Does anyone know if this is the right conclusion