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Question : 64 of 160
Marks:
+1,
-0
Solution:
Given,
x2+y2+sin‌y=4Differentiate both sides w.r.t
x, we get
‌2x+2y⋅‌+cos‌y⋅‌=0⇒‌(2y+cos‌y)=−2x‌‌=‌Again, differentiate both sides w.r.t
x, we get
‌‌=‌(‌)‌=‌| (2y+cos‌y)⋅(−2)−(−2x)⋅(2−sin‌y)⋅ |
| (2y+cos‌y)2 |
‌=‌| −4y−2‌cos‌y+2x⋅(2−sin‌y)⋅() |
| (2y+cos‌y)2 |
‌=‌−4y−2‌cos‌y−| 4x2(2−sin‌y) | | 2y+cos‌y |
|
| (2y+cos‌y)2 |
Put
x=−2 in original equation, we get
‌(−2)2+y2+sin‌y=4⇒‌‌y2+sin‌y=0⇒y=0Now, from Eq. (i), we get
‌‌=‌−4(0)−2‌cos(0)−| 4(−2)2(2−sin‌(0)) | | 2⋅(0)+cos(0) |
|
| (2(0)+cos(0))2 |
‌=‌=‌=−34∴‌‌‌|x=−2‌=−34
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