Question

find dy/dx for y=log (tan2x)

Answer 1

Answer:

$\frac{dy}{dx} = \frac{2}{sin2x cos2x}$

Step-by-step explanation:

y = log (tan 2x)

Applying chain rule:

$\frac{dy}{dx} = \frac{d}{dx} (1st function) . \frac{d}{dx} (2nd function). \frac{d}{dx}(3rd function)$

$\frac{dy}{dx} = \frac{d}{dx} log(tan 2x) . \frac{d}{dx} tan2x. \frac{d}{dx} 2x$

$\frac{dy}{dx} = \frac{1}{tan2x} \times sec^{2}2x \times 2$

$\frac{dy}{dx} = \frac{2sec^{2}2x }{tan2x}$

Now since $sec^{2}x = \frac{1}{cos^{2}x } \ and\ tanx = \frac{sinx}{cosx}$

∴ $\frac{dy}{dx} = \frac{2cos2x }{sin2x cos^22x}$

∴ $\frac{dy}{dx} = \frac{2}{sin2x cos2x}$