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).
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below:
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below:
'इदम्' शब्दस्य स्त्रीलिङ्गे तृतीया-विभक्तौ बहुवचने कि रूपं भवति ?
'कर्तृ' शब्दस्य एकवचनस्य रूपाणि इमानि विभक्त्यनुसारं क्रमेण व्यवस्थापयत ।
(A) कर्त्रा
(B) कर्त्रे
(C) कर्तुः
(D) कर्तारम्
(E) कर्ता
अधोलिखितेषु विकल्पेषु उचिततमम् उत्तरं चिनुत-
प्रथमां सूचीं द्वितीयया सूच्या सह मेलयत ।
सूची-I | सूची-II |
---|---|
(A) षडाननः | (I) यण्-सन्धिः |
(B) यद्यत्र | (II) व्यञ्जन-सन्धिः |
(C) साधुस्तरति | (III) विसर्ग-सन्धिः |
(D) महौषधम् | (IV) वृद्धि-सन्धिः |
अधोलिखितेषु विकल्पेषु उचिततमम् उत्तरं चिनुत -
'उत्+देशः' इत्यत्र सन्धिं कुरुत ।
'दुष्कृतम्' इत्यस्य सन्धि-विच्छेदं कुरुत ।