Solve 2x^2/2-4xquad4.7x^3-x^2/2x^3

$$\frac { 2 x ^ { 2 } } { 2 - 4 x } \quad 4 . \frac { 7 x ^ { 3 } - x ^ { 2 } } { 2 x ^ { 3 } }$$

Evaluate

$$\frac{4x^{2}}{1-2x}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{2x^{2}}{2-4x}.

$$\frac{2x^{2}}{2\left(-2x+1\right)}\times 4$$

Cancel out 2 in both numerator and denominator.

$$\frac{x^{2}}{-2x+1}\times 4$$

Express \frac{x^{2}}{-2x+1}\times 4 as a single fraction.

$$\frac{x^{2}\times 4}{-2x+1}$$

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Quotient

Factor the expressions that are not already factored in \frac{2x^{2}}{2-4x}.

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

Cancel out 2 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x^{2}}{-2x+1}\times 4)$$

Express \frac{x^{2}}{-2x+1}\times 4 as a single fraction.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x^{2}\times 4}{-2x+1})$$

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

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

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}.

$$\frac{\left(-2x^{1}+1\right)\times 2\times 4x^{2-1}-4x^{2}\left(-2\right)x^{1-1}}{\left(-2x^{1}+1\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(-2x^{1}+1\right)\times 8x^{1}-4x^{2}\left(-2\right)x^{0}}{\left(-2x^{1}+1\right)^{2}}$$

Expand using distributive property.

$$\frac{-2x^{1}\times 8x^{1}+8x^{1}-4x^{2}\left(-2\right)x^{0}}{\left(-2x^{1}+1\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{-2\times 8x^{1+1}+8x^{1}-4\left(-2\right)x^{2}}{\left(-2x^{1}+1\right)^{2}}$$

Do the arithmetic.

$$\frac{-16x^{2}+8x^{1}-\left(-8x^{2}\right)}{\left(-2x^{1}+1\right)^{2}}$$

Combine like terms.

$$\frac{\left(-16-\left(-8\right)\right)x^{2}+8x^{1}}{\left(-2x^{1}+1\right)^{2}}$$

Subtract -8 from -16.

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

Factor out 8x.

$$\frac{8x\left(-x^{1}+x^{0}\right)}{\left(-2x^{1}+1\right)^{2}}$$

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

$$\frac{8x\left(-x+x^{0}\right)}{\left(-2x+1\right)^{2}}$$

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

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