If y=sinx.logx,thendy/Dx=
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2
Applying the u.v rule,
d/dx(sinx). logx + sinx. d/dx(logx) =
cosx.logx + sinx. x
d/dx(sinx). logx + sinx. d/dx(logx) =
cosx.logx + sinx. x
Answered by
2
APPY U.V RULE
WE GET,
y = d/DX (sinx.logx) =
sinx.1/x + logx.cosx
sinx/x + logx.cosx
hence this was the answer...
HOPE U GOT IT..MARK ME AS A Brainlest IF U GET IT..THANKS
WE GET,
y = d/DX (sinx.logx) =
sinx.1/x + logx.cosx
sinx/x + logx.cosx
hence this was the answer...
HOPE U GOT IT..MARK ME AS A Brainlest IF U GET IT..THANKS
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