Question

Find dy/dx where y = √log x

Answer 1

Answer:

see the attachment........ ♥

Answer 2

Answer:

$\large \bold\red{\frac{dy}{dx} = \frac{1}{2x \sqrt{ ln(x) } } }$

Step-by-step explanation:

Given,

$y = \sqrt{ ln(x) }$

Now,

To find it's derivative.

Let's assume that,

$ln(x) = t$

Differentiating both sides,

We get,

$= > \frac{1}{x} dx = dt \\ \\ = > \frac{dt}{dx} = \frac{1}{x} \: \: \: \: ..........(i)$

Now,

Substituting the terms,

We get,

$= > y = \sqrt{t}$

Differentiating both sides,

We get,

$= > dy = \frac{1}{2 \sqrt{t} } dt \\ \\ = > \frac{dy}{dt} = \frac{1}{2 \sqrt{t} }$

Substituting the value of t,

We get,

$= > \frac{dy}{dt} = \frac{1}{2 \sqrt{ ln(x) } } \: \: \: \: .........(ii)$

From Equation (i) and (ii),

We get,

$= > \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} \\ \\ = > \frac{dy}{dx} = \frac{1}{2 \sqrt{ ln(x) } } \times \frac{1}{x} \\ \\ = > \large \bold{\frac{dy}{dx} = \frac{1}{2x \sqrt{ ln(x) } } }$