Solve 2xdiv3x^3/x-4

$$2 x \div \frac { 3 x ^ { 3 } } { x - 4 }$$

Evaluate

$$\frac{2\left(x-4\right)}{3x^{2}}$$

Short Solution Steps

Divide 2x by \frac{3x^{3}}{x-4} by multiplying 2x by the reciprocal of \frac{3x^{3}}{x-4}.

$$\frac{2x\left(x-4\right)}{3x^{3}}$$

Cancel out x in both numerator and denominator.

$$\frac{2\left(x-4\right)}{3x^{2}}$$

Use the distributive property to multiply 2 by x-4.

$$\frac{2x-8}{3x^{2}}$$

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Quotient

Divide 2x by \frac{3x^{3}}{x-4} by multiplying 2x by the reciprocal of \frac{3x^{3}}{x-4}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x\left(x-4\right)}{3x^{3}})$$

Cancel out x in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2\left(x-4\right)}{3x^{2}})$$

Use the distributive property to multiply 2 by x-4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x-8}{3x^{2}})$$

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.

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

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}.

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

Do the arithmetic.

$$\frac{3x^{2}\times 2x^{0}-\left(2x^{1}-8\right)\times 6x^{1}}{\left(3x^{2}\right)^{2}}$$

Expand using distributive property.

$$\frac{3x^{2}\times 2x^{0}-\left(2x^{1}\times 6x^{1}-8\times 6x^{1}\right)}{\left(3x^{2}\right)^{2}}$$

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

$$\frac{3\times 2x^{2}-\left(2\times 6x^{1+1}-8\times 6x^{1}\right)}{\left(3x^{2}\right)^{2}}$$

Do the arithmetic.

$$\frac{6x^{2}-\left(12x^{2}-48x^{1}\right)}{\left(3x^{2}\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{6x^{2}-12x^{2}-\left(-48x^{1}\right)}{\left(3x^{2}\right)^{2}}$$

Combine like terms.

$$\frac{\left(6-12\right)x^{2}-\left(-48x^{1}\right)}{\left(3x^{2}\right)^{2}}$$

Subtract 12 from 6.

$$\frac{-6x^{2}-\left(-48x^{1}\right)}{\left(3x^{2}\right)^{2}}$$

Factor out 6x.

$$\frac{6x\left(-x^{1}-\left(-8x^{0}\right)\right)}{\left(3x^{2}\right)^{2}}$$

To raise the product of two or more numbers to a power, raise each number to the power and take their product.

$$\frac{6x\left(-x^{1}-\left(-8x^{0}\right)\right)}{3^{2}\left(x^{2}\right)^{2}}$$

Raise 3 to the power 2.

$$\frac{6x\left(-x^{1}-\left(-8x^{0}\right)\right)}{9\left(x^{2}\right)^{2}}$$

To raise a power to another power, multiply the exponents.

$$\frac{6x\left(-x^{1}-\left(-8x^{0}\right)\right)}{9x^{2\times 2}}$$

Multiply 2 times 2.

$$\frac{6x\left(-x^{1}-\left(-8x^{0}\right)\right)}{9x^{4}}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{6\left(-x^{1}-\left(-8x^{0}\right)\right)}{9x^{4-1}}$$

Subtract 1 from 4.

$$\frac{6\left(-x^{1}-\left(-8x^{0}\right)\right)}{9x^{3}}$$

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

$$\frac{6\left(-x-\left(-8x^{0}\right)\right)}{9x^{3}}$$

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

$$\frac{6\left(-x-\left(-8\right)\right)}{9x^{3}}$$