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Answer 1 · 2017-10-15T14:41:17

$4 x$ $4x$

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

...so you can rewrite as:

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

...and I think this simplifies to:

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

And the derivative of this would b:

$- 4 x$ $-4x$

GOOD LUCK

Answer 2 · 2017-10-15T14:49:59

Given: $\sqrt{1} - x^{2} - \frac{x}{\sqrt{1}} - x^{2} + x$ $sqrt1-x^2-x/sqrt1-x^2+x$

Use the substitution, $\sqrt{1} = 1$ $sqrt1 = 1$ :

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

Combine like terms:

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

Differentiate:

$\frac{d (1 - 2 x^{2})}{d x} = \frac{d (1)}{d x} - \frac{d (2 x^{2})}{d x}$ $(d(1-2x^2))/dx = (d(1))/dx -(d(2x^2))/dx$

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

$\frac{d (1 - 2 x^{2})}{d x} = - \frac{d (2 x^{2})}{d x}$ $(d(1-2x^2))/dx = - (d(2x^2))/dx$

Use the linear property of the derivative:

$\frac{d (1 - 2 x^{2})}{d x} = - 2 \frac{d (x^{2})}{d x}$ $(d(1-2x^2))/dx = - 2(d(x^2))/dx$

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

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