Solve fff:x^4-2x^2-8=

$$f f f : x ^ { 4 } - 2 x ^ { 2 } - 8 =$$

Evaluate

$$-2x^{2}-8+\frac{f^{3}}{x^{4}}$$

Short Solution Steps

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

$$\frac{f^{2}f}{x^{4}}-2x^{2}-8$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{f^{3}}{x^{4}}-2x^{2}-8$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x^{2}-8 times \frac{x^{4}}{x^{4}}.

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

Since \frac{f^{3}}{x^{4}} and \frac{\left(-2x^{2}-8\right)x^{4}}{x^{4}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in f^{3}+\left(-2x^{2}-8\right)x^{4}.

$$\frac{f^{3}-2x^{6}-8x^{4}}{x^{4}}$$

Differentiate w.r.t. x

$$-4x-\frac{4f^{3}}{x^{5}}$$