Question:

Let \( y=\sin(\cos(x^2)) \). Find \( \frac{dy}{dx} \) at \( x=\frac{\sqrt{\pi}}{2} \).

Show Hint

When differentiating nested functions, apply the chain rule from the outside in. Differentiate the outermost function, keeping the inside intact, then multiply by the derivative of the next function inside, and so on, until you reach the innermost variable. Be careful with signs, especially when differentiating cosine.
Updated On: Sep 9, 2025
  • $-\frac{\sqrt{\pi}}{2} \cos(\frac{1}{\sqrt{2}})$
  • $-\sqrt{\pi} \cos(\frac{1}{\sqrt{2}})$
  • $-\frac{\sqrt{\pi}}{2} \sin(\frac{1}{\sqrt{2}})$
  • $\sqrt{\frac{\pi}{2}} \sin(\frac{1}{\sqrt{2}})$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:

We need to find the derivative of a composite function, which requires the use of the chain rule multiple times. After finding the derivative, we will evaluate it at the specified point.

Step 2: Key Formula or Approach:

The chain rule states that if \( y = f(g(h(x))) \), then: \[ \frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \] We have \( y = \sin(u) \), \( u = \cos(v) \), and \( v = x^2 \).

Step 3: Detailed Explanation:

The function is \( y = \sin(\cos(x^2)) \). Let's apply the chain rule to find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{d}{dx} \sin(\cos(x^2)) \] \[ = \cos(\cos(x^2)) \cdot \frac{d}{dx}(\cos(x^2)) \] \[ = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot \frac{d}{dx}(x^2) \] \[ = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot (2x) \] \[ \frac{dy}{dx} = -2x \sin(x^2) \cos(\cos(x^2)) \] Now, we evaluate this derivative at \( x = \frac{\sqrt{\pi}}{2} \). First, let's find the values of the terms involving \( x \): \[ x^2 = \left(\frac{\sqrt{\pi}}{2}\right)^2 = \frac{\pi}{4} \] So we have: \[ \sin(x^2) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \] \[ \cos(x^2) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \] Substitute these values back into the expression for the derivative: \[ \frac{dy}{dx} \bigg|_{x=\frac{\sqrt{\pi}}{2}} = -2\left(\frac{\sqrt{\pi}}{2}\right) \cdot \sin\left(\frac{\pi}{4}\right) \cdot \cos\left(\cos\left(\frac{\pi}{4}\right)\right) \] \[ = -\sqrt{\pi} \cdot \left(\frac{1}{\sqrt{2}}\right) \cdot \cos\left(\frac{1}{\sqrt{2}}\right) \] \[ = -\frac{\sqrt{\pi}}{\sqrt{2}} \cos\left(\frac{1}{\sqrt{2}}\right) \] This can also be written as \( -\sqrt{\frac{\pi}{2}} \cos\left(\frac{1}{\sqrt{2}}\right) \).

Step 4: Final Answer:

The value of the derivative at \( x = \frac{\sqrt{\pi}}{2} \) is \( -\frac{\sqrt{\pi}}{\sqrt{2}} \cos\left(\frac{1}{\sqrt{2}}\right) \). This matches option (A).

Was this answer helpful?
0
0

Top Questions on Continuity and differentiability

View More Questions

Questions Asked in CUET exam

View More Questions