# Question #9cdcc

Oct 15, 2017

$4 x$

#### Explanation:

well, the radical signs are unnecessary. $\sqrt{1} = 1$

...so you can rewrite as:

$1 - {x}^{2} - x - {x}^{2} + x$

...and I think this simplifies to:

$1 - 2 {x}^{2}$

And the derivative of this would b:

$- 4 x$

GOOD LUCK

Oct 15, 2017

Given: $\sqrt{1} - {x}^{2} - \frac{x}{\sqrt{1}} - {x}^{2} + x$

Use the substitution, $\sqrt{1} = 1$:

$1 - {x}^{2} - x - {x}^{2} + x$

Combine like terms:

$1 - 2 {x}^{2}$

Differentiate:

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

The first term is 0 because the derivative of a constant is 0:

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

Use the linear property of the derivative:

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

Use the power rule, $\frac{d \left({x}^{n}\right)}{\mathrm{dx}} = n {x}^{n - 1}$:

$\frac{d \left(1 - 2 {x}^{2}\right)}{\mathrm{dx}} = - 4 x$