Differentiate:
$tan (2 x + 1)$

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Solution

$y = tan (2 x + 1)$

$\frac{d y}{d x} = {sec}^{2} (2 x + 1) \frac{d}{d x} (2 x + 1)$

$\frac{d y}{d x} = {sec}^{2} (2 x + 1) (2)$

$∴ \frac{d y}{d x} = 2 {sec}^{2} (2 x + 1)$

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Similar questions

Differentiate each of the following from first principles :

$(i) t a n^{2} x$

$(i i) t a n (2 x + 1)$

$(i i i) t a n 2 x$

$(i i v) \sqrt{t a n x}$

\int

\frac{1 + t a n 2 x}{1 - t a n 2 x}

\int \frac{1 + {tan}^{2} x}{\sqrt{{tan}^{2} x + 3}}

cos 2 x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x}

tan 2 x = \frac{2 tan x}{1 - {tan}^{2} x}

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