Question

x + logex.
Find dy , if y = 1
dx​

Answer 1

Answer:

As y = 1/x (log x)

Then dy/dx = d/dx (1/x)•log x + 1/x • d/dx (log x)

Also dy/dx = -1/x^2 •log x + 1/x • 1/x

So dy/dx = -log x/x^2 + 1/x^2

Step-by-step explanation:

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