# Calculating Derivative – A quotient of a polynom and square root – Exercise 6267

Exercise

Find the derivative of the following function:

$$f(x)=\frac{x^4-x^2+2}{x\sqrt{x}}$$

$$f'(x)=\frac{5x^4-x^2+6}{2x^2\sqrt{x}}$$

Solution

We simplify the function before differentiating:

$$f(x)=\frac{x^4-x^2+2}{x\sqrt{x}}=$$

$$f(x)=\frac{x^4-x^2+2}{x^{\frac{3}{2}}}$$

Using Derivative formulas and the quotient rule in Derivative Rules, we get the derivative:

$$f'(x)=\frac{(4x^3-2x)\cdot x\sqrt{x}-(x^4-x^2+2)\cdot \frac{3}{2}\sqrt{x}}{{(x\sqrt{x})}^2}=$$

One can simplify the derivative:

$$=\frac{4x^4\sqrt{x}-2x^2\sqrt{x}-\frac{3}{2}x^4\sqrt{x}+\frac{3}{2}x^2\sqrt{x}+3\sqrt{x}}{x^2\cdot x}=$$

$$=\frac{\frac{5}{2}x^4\sqrt{x}-\frac{1}{2}x^2\sqrt{x}+3\sqrt{x}}{x^3}=$$

$$=\frac{5x^4\sqrt{x}-x^2\sqrt{x}+6\sqrt{x}}{2x^3}=$$

$$=\frac{5x^4-x^2+6}{2x^2\sqrt{x}}$$

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