Question
If sin (x+y)=loge(x+y),thendydx is equal to
Answer: Option D
:
D
∵sin(x+y)=loge(x+y),
cos(x+y)(1+dydx)=1(x+y)(1+dydx)
⇒(1+dydx)(cos(x+y)−1x+y)=0
∵cos(x+y)−1(x+y)≠ 0
∴1+dydx=0
∴dydx=−1
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:
D
∵sin(x+y)=loge(x+y),
cos(x+y)(1+dydx)=1(x+y)(1+dydx)
⇒(1+dydx)(cos(x+y)−1x+y)=0
∵cos(x+y)−1(x+y)≠ 0
∴1+dydx=0
∴dydx=−1
Was this answer helpful ?
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