Question

The derivative of the function $f(x) = \sin(x^2)$ at $x = \sqrt{\pi}$ is

• A. $2\sqrt{\pi} \cos(\pi)$
• B. $\sqrt{\pi} \sin(\pi)$
• C. $-\sqrt{\pi} \cos(\pi)$
• D. $2\pi \sin(\pi)$

Hint:

When differentiating composite functions like $\sin(x^2)$, always apply the chain rule by differentiating the outer function and multiplying by the derivative of the inner function.

Answer 1

The Correct Option is A

To find the derivative of $f(x) = \sin(x^2)$, we use the chain rule. Let $u = x^2$, so $f(x) = \sin(u)$ and $u = x^2$. Then, $\frac{df}{du} = \cos(u)$ and $\frac{du}{dx} = 2x$. Thus, the derivative is: \[ \begin{align} f'(x) = \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \cos(x^2) \cdot 2x \] Now, evaluate at $x = \sqrt{\pi}$: \[ \begin{align} f'(\sqrt{\pi}) = 2\sqrt{\pi} \cdot \cos((\sqrt{\pi})^2) = 2\sqrt{\pi} \cdot \cos(\pi) \] Since $\cos(\pi) = -1$, we have: \[ \begin{align} f'(\sqrt{\pi}) = 2\sqrt{\pi} \cdot (-1) = -2\sqrt{\pi} \] However, checking the options, we notice they imply a possible adjustment in interpretation. The correct form matches option (1) when considering the magnitude or a potential variation in the problem setup. The intended answer based on options is $2\sqrt{\pi} \cos(\pi)$, which aligns with the given correct answer. Thus, the correct answer is $2\sqrt{\pi} \cos(\pi)$.