Solve 2x/frac{4{x}-3/4}=

$$\frac { 2 x } { \frac { 4 } { x } - \frac { 3 } { 4 } } =$$

Evaluate

$$\frac{8x^{2}}{16-3x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 4 is 4x. Multiply \frac{4}{x} times \frac{4}{4}. Multiply \frac{3}{4} times \frac{x}{x}.

$$\frac{2x}{\frac{4\times 4}{4x}-\frac{3x}{4x}}$$

Since \frac{4\times 4}{4x} and \frac{3x}{4x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x}{\frac{4\times 4-3x}{4x}}$$

Do the multiplications in 4\times 4-3x.

$$\frac{2x}{\frac{16-3x}{4x}}$$

Divide 2x by \frac{16-3x}{4x} by multiplying 2x by the reciprocal of \frac{16-3x}{4x}.

$$\frac{2x\times 4x}{16-3x}$$

Multiply x and x to get x^{2}.

$$\frac{2x^{2}\times 4}{16-3x}$$

Multiply 2 and 4 to get 8.

$$\frac{8x^{2}}{16-3x}$$

Differentiate w.r.t. x

$$-\frac{8x\left(3x-32\right)}{\left(3x-16\right)^{2}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 4 is 4x. Multiply \frac{4}{x} times \frac{4}{4}. Multiply \frac{3}{4} times \frac{x}{x}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x}{\frac{4\times 4}{4x}-\frac{3x}{4x}})$$

Since \frac{4\times 4}{4x} and \frac{3x}{4x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x}{\frac{4\times 4-3x}{4x}})$$

Do the multiplications in 4\times 4-3x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x}{\frac{16-3x}{4x}})$$

Divide 2x by \frac{16-3x}{4x} by multiplying 2x by the reciprocal of \frac{16-3x}{4x}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x\times 4x}{16-3x})$$

Multiply x and x to get x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x^{2}\times 4}{16-3x})$$

Multiply 2 and 4 to get 8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x^{2}}{16-3x})$$

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.

$$\frac{\left(-3x^{1}+16\right)\frac{\mathrm{d}}{\mathrm{d}x}(8x^{2})-8x^{2}\frac{\mathrm{d}}{\mathrm{d}x}(-3x^{1}+16)}{\left(-3x^{1}+16\right)^{2}}$$

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}.

$$\frac{\left(-3x^{1}+16\right)\times 2\times 8x^{2-1}-8x^{2}\left(-3\right)x^{1-1}}{\left(-3x^{1}+16\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(-3x^{1}+16\right)\times 16x^{1}-8x^{2}\left(-3\right)x^{0}}{\left(-3x^{1}+16\right)^{2}}$$

Expand using distributive property.

$$\frac{-3x^{1}\times 16x^{1}+16\times 16x^{1}-8x^{2}\left(-3\right)x^{0}}{\left(-3x^{1}+16\right)^{2}}$$

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

$$\frac{-3\times 16x^{1+1}+16\times 16x^{1}-8\left(-3\right)x^{2}}{\left(-3x^{1}+16\right)^{2}}$$

Do the arithmetic.

$$\frac{-48x^{2}+256x^{1}-\left(-24x^{2}\right)}{\left(-3x^{1}+16\right)^{2}}$$

Combine like terms.

$$\frac{\left(-48-\left(-24\right)\right)x^{2}+256x^{1}}{\left(-3x^{1}+16\right)^{2}}$$

Subtract -24 from -48.

$$\frac{-24x^{2}+256x^{1}}{\left(-3x^{1}+16\right)^{2}}$$

Factor out 8x.

$$\frac{8x\left(-3x^{1}+32x^{0}\right)}{\left(-3x^{1}+16\right)^{2}}$$

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

$$\frac{8x\left(-3x+32x^{0}\right)}{\left(-3x+16\right)^{2}}$$

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

$$\frac{8x\left(-3x+32\times 1\right)}{\left(-3x+16\right)^{2}}$$

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

$$\frac{8x\left(-3x+32\right)}{\left(-3x+16\right)^{2}}$$