Differentiate the following functions with respect to x :

$\frac{x}{1 + t a n x}$

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Solution

Let y = $\frac{x}{1 + t a n x}$

Differentiating y w.r.t. x, we get

$\frac{d y}{d x} = \frac{(1 + t a n x) \frac{d}{d x} (x) - x \frac{d}{d x} (1 + t a n x)}{(1 + t a n x)^{2}}$

$= \frac{(1 + t a n x) (1) - x (0 + s e c^{2} x)}{(1 + t a n x)^{2}}$

$= \frac{1 + t a n x - x s e c^{2} x}{(1 + t a n x)^{2}}$

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