Question

If y = (cosx²)², then \(\frac{dy}{dx}\) is equal to

• A. -4x sin2x²
• B. -x sinx²
• C. -2x sin2x²
• D. -x cos2x²

Answer 1

The Correct Option is C

Given: \( y = (\cos(x^2))^2 \)
We need to find \( \frac{dy}{dx} \).

Step 1: Simplifying the function
\( y = \cos^2(x^2) \) can be written as:
\[ y = (\cos(u))^2 \quad \text{where} \quad u = x^2 \]
Step 2: Applying chain rule
Using the chain rule, we differentiate \( y = (\cos(u))^2 \). First, differentiate the outer function: \[ \frac{d}{du} \left[ (\cos(u))^2 \right] = 2\cos(u) \cdot (-\sin(u)) \] So, we get: \[ \frac{dy}{du} = -2\cos(u)\sin(u) \] Next, differentiate \( u = x^2 \) with respect to \( x \): \[ \frac{du}{dx} = 2x \]
Step 3: Combining the results
Now applying the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = -2\cos(x^2) \sin(x^2) \times 2x \] Simplifying: \[ \frac{dy}{dx} = -4x \cos(x^2) \sin(x^2) \]
Step 4: Final Answer
The expression simplifies to \( -2x \sin(2x^2) \) (since \( 2\sin(A)\cos(A) = \sin(2A) \)).

The correct answer is (C).

Answer 2

Given \( y = (\cos x^2)^2 \), we want to find \( \frac{dy}{dx} \).

We can use the chain rule to differentiate this function. The chain rule states that if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

Let \( u = \cos x^2 \), so \( y = u^2 \). Then \( \frac{dy}{du} = 2u \).

Now we need to find \( \frac{du}{dx} \). Since \( u = \cos x^2 \), let \( v = x^2 \), so \( u = \cos v \). Then \( \frac{du}{dv} = -\sin v \).

And \( \frac{dv}{dx} = \frac{d}{dx}(x^2) = 2x \).

Now we can find \( \frac{du}{dx} \) using the chain rule: \( \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = (-\sin v)(2x) = -2x \sin x^2 \).

Finally, we can find \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (2u)(-2x \sin x^2) = 2(\cos x^2)(-2x \sin x^2) = -4x \cos x^2 \sin x^2 \).

We can simplify this using the trigonometric identity \( 2 \sin \theta \cos \theta = \sin 2\theta \). So, \( \cos x^2 \sin x^2 = \frac{1}{2} \sin(2x^2) \).

Therefore, \( \frac{dy}{dx} = -4x \left( \frac{1}{2} \sin(2x^2) \right) = -2x \sin(2x^2) \).