Question

If $ f(x) = e^{2x} \sin x $, find $ f'(x) $.

• A. \( e^{2x}(2\sin x + \cos x) \)
• B. \( e^{2x}(2\sin x - \cos x) \)
• C. \( e^{2x}(2\cos x + \sin x) \)
• D. \( e^{2x}(\sin x - 2\cos x) \)

Hint:

For differentiation of a product of two functions, use the product rule: \( (uv)' = u'v + uv' \).

Answer 1

The Correct Option is A

To find the derivative \( f'(x) \) of the function \( f(x) = e^{2x} \sin x \), we apply the product rule of differentiation. The product rule states that if you have a function \( f(x) = u(x) v(x) \), then the derivative \( f'(x) = u'(x) v(x) + u(x) v'(x) \).

Here, let \( u(x) = e^{2x} \) and \( v(x) = \sin x \). We need to find \( u'(x) \) and \( v'(x) \):

\( u'(x) = \frac{d}{dx}[e^{2x}] = 2e^{2x} \) (using the chain rule)
\( v'(x) = \frac{d}{dx}[\sin x] = \cos x \)

Now, apply the product rule:

\( f'(x) = u'(x) v(x) + u(x) v'(x) \)

\( = (2e^{2x})(\sin x) + (e^{2x})(\cos x) \)

\( = e^{2x}(2\sin x) + e^{2x}(\cos x) \)

\( = e^{2x}(2\sin x + \cos x) \)

Thus, the derivative is \( f'(x) = e^{2x}(2\sin x + \cos x) \).