Saurabh Koranglekar
Last Activity: 6 Years ago
To find the derivative of the function \( Y = e^{\sin^2 x + \sin^4 x + \sin^6 x + \ldots} \), we first need to understand the infinite series in the exponent. Let's break it down step by step.
Understanding the Infinite Series
The expression in the exponent is a series of sine terms raised to even powers. Specifically, we can express it as:
- \( S = \sin^2 x + \sin^4 x + \sin^6 x + \ldots \)
This series can be recognized as a geometric series where the first term \( a = \sin^2 x \) and the common ratio \( r = \sin^2 x \) as well. The sum of an infinite geometric series is given by the formula:
S = a / (1 - r) provided that \( |r| < 1 \).
Finding the Sum of the Series
Applying the formula, we get:
S = \frac{\sin^2 x}{1 - \sin^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x.
Thus, we can rewrite \( Y \) as:
Y = e^{\tan^2 x}.
Calculating the Derivative
To find the derivative \( \frac{dY}{dx} \), we will use the chain rule. The chain rule states that if you have a function \( Y = e^{u} \), where \( u = \tan^2 x \), then:
\(\frac{dY}{dx} = e^{u} \cdot \frac{du}{dx}.\)
Finding \( \frac{du}{dx} \)
Now, we need to compute \( \frac{du}{dx} \) where \( u = \tan^2 x \). Using the chain rule again, we have:
\(\frac{du}{dx} = 2\tan x \cdot \sec^2 x.\)
Putting It All Together
Now that we have \( \frac{du}{dx} \), we can substitute back into our expression for \( \frac{dY}{dx} \):
\(\frac{dY}{dx} = e^{\tan^2 x} \cdot (2\tan x \cdot \sec^2 x).\)
To summarize, the derivative of \( Y \) with respect to \( x \) is:
\(\frac{dY}{dx} = e^{\tan^2 x} \cdot 2\tan x \cdot \sec^2 x.\)
Final Thoughts
This result gives us valuable insight into how the function behaves as \( x \) changes. The presence of \( e^{\tan^2 x} \) suggests that the function grows rapidly based on the input from the sine function's behavior. Understanding how derivatives work in this context can be very helpful in many areas of calculus, especially when dealing with exponential growth and trigonometric functions.