Solve 1^-1(1/1-x+x^2)

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

Evaluate

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

Solution Steps

Calculate 1 to the power of -1 and get 1.

$$1\times \frac{1}{1-x+x^{2}}$$

Express 1\times \frac{1}{1-x+x^{2}} as a single fraction.

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

Differentiate w.r.t. x

$$-\frac{2x-1}{\left(x^{2}-x+1\right)^{2}}$$

Steps Using Chain Rule

Calculate 1 to the power of -1 and get 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(1\times \frac{1}{1-x+x^{2}})$$

Express 1\times \frac{1}{1-x+x^{2}} as a single fraction.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{1-x+x^{2}})$$

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

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

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$-\left(-x^{1}+x^{2}+1\right)^{-2}\left(-x^{1-1}+2x^{2-1}\right)$$

Simplify.

$$\left(-x^{1}+x^{2}+1\right)^{-2}\left(x^{0}-2x^{1}\right)$$

For any term t, t^{1}=t.

$$\left(-x+x^{2}+1\right)^{-2}\left(x^{0}-2x\right)$$

For any term t except 0, t^{0}=1.

$$\left(-x+x^{2}+1\right)^{-2}\left(1-2x\right)$$