Question

(x-1/x)(x square + 1/x square ) differentiate it w.r.t x by dis formula u(du/dx)+ v (du/dx)

Answer 1

Answer on Question #47270 – Math – Calculus

$(x - 1 / x)(x \text{ square} + 1 / x \text{ square})$ differentiate it w.r.t $x$ by dis formula $u(du/dx) + v(du/dx)$

Solution:

$$y = \left(x - \frac {1}{x}\right) \left(x ^ {2} + \frac {1}{x ^ {2}}\right)$$

Derivative product rule:

$$(f g) ^ {\prime} = f ^ {\prime} g + f g ^ {\prime}$$

In our case:

$$\begin{array}{l} f = \left(x - \frac {1}{x}\right); g = \left(x ^ {2} + \frac {1}{x ^ {2}}\right) \\ y ^ {\prime} = (f g) ^ {\prime} = \left(1 + \frac {1}{x ^ {2}}\right) \left(x ^ {2} + \frac {1}{x ^ {2}}\right) + \left(x - \frac {1}{x}\right) \left(2 x - \frac {2}{x ^ {3}}\right) = \\ = \frac {3}{x ^ {4}} + 3 x ^ {2} - \frac {1}{x ^ {2}} - 1 \\ \end{array}$$

Answer: $y' = \frac{3}{x^4} + 3x^2 - \frac{1}{x^2} - 1$.

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