Question

if ( y=arctan left(e^{2 x}
ight) ), then ( \frac{d y}{d x}= )

Explanation:

Step1: Let \(u = e^{2x}\)

The function \(y=\arctan(u)\). By the chain - rule, \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
The derivative of \(y = \arctan(u)\) with respect to \(u\) is \(\frac{dy}{du}=\frac{1}{1 + u^{2}}\).

Step2: Find \(\frac{du}{dx}\)

Since \(u = e^{2x}\), using the chain - rule for \(y = e^{v}\) where \(v = 2x\). The derivative of \(e^{v}\) with respect to \(v\) is \(e^{v}\), and the derivative of \(v=2x\) with respect to \(x\) is \(2\). So \(\frac{du}{dx}=e^{2x}\cdot2 = 2e^{2x}\).

Step3: Substitute \(u\) back and calculate \(\frac{dy}{dx}\)

Substitute \(u = e^{2x}\) into \(\frac{dy}{du}\), we get \(\frac{dy}{du}=\frac{1}{1+(e^{2x})^{2}}=\frac{1}{1 + e^{4x}}\).
Then \(\frac{dy}{dx}=\frac{1}{1 + e^{4x}}\cdot2e^{2x}=\frac{2e^{2x}}{1 + e^{4x}}\).

Answer:

\(\frac{2e^{2x}}{1 + e^{4x}}\)