Solve d/dxleft(x^3-8/x^2right)

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

Evaluate

$$1+\frac{16}{x^{3}}$$

Steps Using Derivative Rule for Quotient

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.

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

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}.

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

Do the arithmetic.

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

Expand using distributive property.

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

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

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

Do the arithmetic.

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

Remove unnecessary parentheses.

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

Combine like terms.

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

Subtract 2 from 3.

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

Factor out x.

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

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

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

Multiply 2 times 2.

$$\frac{x\left(x^{3}-\left(-16x^{0}\right)\right)}{x^{4}}$$

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

$$\frac{x^{3}-\left(-16x^{0}\right)}{x^{4-1}}$$

Subtract 1 from 4.

$$\frac{x^{3}-\left(-16x^{0}\right)}{x^{3}}$$

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

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

Differentiate w.r.t. x

$$-\frac{48}{x^{4}}$$