Solve d/dxleft(8x+1/2x^2right)

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

Evaluate

$$8-\frac{1}{x^{3}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Multiply 8x times \frac{2x^{2}}{2x^{2}}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x\times 2x^{2}}{2x^{2}}+\frac{1}{2x^{2}})$$

Since \frac{8x\times 2x^{2}}{2x^{2}} and \frac{1}{2x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8x\times 2x^{2}+1}{2x^{2}})$$

Do the multiplications in 8x\times 2x^{2}+1.

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

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.

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

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}.

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

Do the arithmetic.

$$\frac{2x^{2}\times 48x^{2}-\left(16x^{3}+1\right)\times 4x^{1}}{\left(2x^{2}\right)^{2}}$$

Expand using distributive property.

$$\frac{2x^{2}\times 48x^{2}-\left(16x^{3}\times 4x^{1}+4x^{1}\right)}{\left(2x^{2}\right)^{2}}$$

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

$$\frac{2\times 48x^{2+2}-\left(16\times 4x^{3+1}+4x^{1}\right)}{\left(2x^{2}\right)^{2}}$$

Do the arithmetic.

$$\frac{96x^{4}-\left(64x^{4}+4x^{1}\right)}{\left(2x^{2}\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{96x^{4}-64x^{4}-4x^{1}}{\left(2x^{2}\right)^{2}}$$

Combine like terms.

$$\frac{\left(96-64\right)x^{4}-4x^{1}}{\left(2x^{2}\right)^{2}}$$

Subtract 64 from 96.

$$\frac{32x^{4}-4x^{1}}{\left(2x^{2}\right)^{2}}$$

Factor out 4x.

$$\frac{4x\left(8x^{3}-x^{0}\right)}{\left(2x^{2}\right)^{2}}$$

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

$$\frac{4x\left(8x^{3}-x^{0}\right)}{2^{2}\left(x^{2}\right)^{2}}$$

Raise 2 to the power 2.

$$\frac{4x\left(8x^{3}-x^{0}\right)}{4\left(x^{2}\right)^{2}}$$

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

$$\frac{4x\left(8x^{3}-x^{0}\right)}{4x^{2\times 2}}$$

Multiply 2 times 2.

$$\frac{4x\left(8x^{3}-x^{0}\right)}{4x^{4}}$$

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

$$\frac{4\left(8x^{3}-x^{0}\right)}{4x^{4-1}}$$

Subtract 1 from 4.

$$\frac{4\left(8x^{3}-x^{0}\right)}{4x^{3}}$$

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

$$\frac{4\left(8x^{3}-1\right)}{4x^{3}}$$

Differentiate w.r.t. x

$$\frac{3}{x^{4}}$$