Answer:
[tex]\frac{dy}{dx} = \frac{2}{sin2x cos2x}[/tex]
Step-by-step explanation:
y = log (tan 2x)
Applying chain rule:
[tex]\frac{dy}{dx} = \frac{d}{dx} (1st function) . \frac{d}{dx} (2nd function). \frac{d}{dx}(3rd function)[/tex]
[tex]\frac{dy}{dx} = \frac{d}{dx} log(tan 2x) . \frac{d}{dx} tan2x. \frac{d}{dx} 2x[/tex]
[tex]\frac{dy}{dx} = \frac{1}{tan2x} \times sec^{2}2x \times 2[/tex]
[tex]\frac{dy}{dx} = \frac{2sec^{2}2x }{tan2x}[/tex]
Now since [tex]sec^{2}x = \frac{1}{cos^{2}x } \ and\ tanx = \frac{sinx}{cosx}[/tex]
∴ [tex]\frac{dy}{dx} = \frac{2cos2x }{sin2x cos^22x}[/tex]
∴ [tex]\frac{dy}{dx} = \frac{2}{sin2x cos2x}[/tex]
