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AfilCa [17]
3 years ago
7

Find dy/dx for y=log (tan2x)

Mathematics
1 answer:
nekit [7.7K]3 years ago
7 0

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}

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