Solve (x^4+frac{97/sqrt{32})}{-3}

$$\frac { ( x ^ { 4 } + \frac { 97 } { \sqrt { 32 } } ) } { - 3 }$$

Evaluate

$$-\frac{x^{4}}{3}-\frac{97\sqrt{2}}{24}$$

Short Solution Steps

Multiply both numerator and denominator by -1.

$$\frac{-x^{4}-\frac{97}{\sqrt{32}}}{3}$$

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$\frac{-x^{4}-\frac{97}{4\sqrt{2}}}{3}$$

Rationalize the denominator of \frac{97}{4\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{-x^{4}-\frac{97\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}}{3}$$

The square of \sqrt{2} is 2.

$$\frac{-x^{4}-\frac{97\sqrt{2}}{4\times 2}}{3}$$

Multiply 4 and 2 to get 8.

$$\frac{-x^{4}-\frac{97\sqrt{2}}{8}}{3}$$

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

$$\frac{-\frac{8x^{4}}{8}-\frac{97\sqrt{2}}{8}}{3}$$

Since -\frac{8x^{4}}{8} and \frac{97\sqrt{2}}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{-8x^{4}-97\sqrt{2}}{8}}{3}$$

Express \frac{\frac{-8x^{4}-97\sqrt{2}}{8}}{3} as a single fraction.

$$\frac{-8x^{4}-97\sqrt{2}}{8\times 3}$$

Multiply 8 and 3 to get 24.

$$\frac{-8x^{4}-97\sqrt{2}}{24}$$

Differentiate w.r.t. x

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