if y=log 3x+log(2/x) then find dy/dx
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Answer:1/x for all the three cases when differentiating w.r.t. x
Soln. We know d(logx)/dx =1/x.
Now, d(log2x)/dx =(1/2x)*d(2x)/dx => 1*2/2x=>1/x {first differntiating log2x , then derivative of 2x applying chain rule}
Similarly, d(log3x)/dx = 3/3x=1/x and d(log4x)/dx = 1*4/4x = 1/x . Therefore the derivative of log2x , log3x and log4x is 1/x
Step-by-step explanation:
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