Question

find the first derivative of y=√logx
​

Answer 1

Answer:

y=

logx+

logx+

logx+......∞

.......(1).

Squaring both sides we've,

y

2

=logx+

logx+

logx+

logx+......

or, y

2

=logx+y [ Using (1)]

Now differentiating both sides with respect to x we get,

(2y−1)

dx

d dy = x1

or,

dxdd=

x(2y−1)

1

.

Answer 2

Answer:

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