Question

If $y = \cos x$ then $\frac{d^2y}{dx^2} =$

• A. $-\cos x$
• B. $-\sin x$
• C. $\cos x$
• D. $\sin x$

Hint:

The derivatives of sine and cosine follow a four-step cycle: $\cos x \to -\sin x \to -\cos x \to \sin x \to \cos x$. Knowing this cycle allows you to find any $n^{th}$ order derivative quickly!

Answer 1

The Correct Option is A

Step 1: Find the first derivative ($\frac{dy}{dx}$): The derivative of $\cos x$ with respect to $x$ is: $$\frac{dy}{dx} = -\sin x$$

Step 2: Find the second derivative ($\frac{d^2y}{dx^2}$): Differentiate the first derivative again with respect to $x$: $$\frac{d^2y}{dx^2} = \frac{d}{dx}(-\sin x)$$ The derivative of $\sin x$ is $\cos x$. Therefore: $$\frac{d^2y}{dx^2} = -(\cos x) = -\cos x$$ This result shows that differentiating $\cos x$ twice returns the negative of the original function.