Solve (1/1-2x)^prime

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

Evaluate

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

Steps Using Chain Rule

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).

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

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}.

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

Simplify.

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

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

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

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

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

For any term t, t\times 1=t and 1t=t.

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

Differentiate w.r.t. x

$$\frac{8}{\left(1-2x\right)^{3}}$$