Question

differentiation of √x - 1/√x​

Answer 1

Step-by-step explanation:

Let y=√x+1/√x eqn1

For x+∆x, y+∆y=√(x+∆x)+1/(√(x+∆x).

=((x+∆x)+1)((x+∆x)^-0.5)

=(x+∆x+1)(x^-0.5)((1+(∆x/x))^-0.5)

Using binomial theorem,

(x+∆x+1)(x^-0.5)(1-(0.5(∆x/x)))

Hence, y+∆y = (x+∆x+1)((1/√x)-(∆x/2x√x)) eqn2

eqn2-eqn1 gives,

∆y= √x +(∆x/√x) +1/√x -(x∆x/2x√x) - ((∆x)^2/2x√x) - (∆x/2x√x) -√x -(1/√x)

Hence,

∆y = (∆x/√x) - (∆x/2√x) -((∆x)^2/2x√x) - (∆x/2x√x)

∆y/∆x = 1/2√x - (1/2x√x) - (∆x/2x√x)

Applying limit as ∆x -> 0 and Simplifying further,

dy/dx = 1/2√x - (1/2x√x)

Hope this helps.

Answer 2

Answer:

Let y=√x+1/√x eqn1

For x+∆x, y+∆y=√(x+∆x)+1/(√(x+∆x).

=((x+∆x)+1)((x+∆x)^-0.5)

=(x+∆x+1)(x^-0.5)((1+(∆x/x))^-0.5)

Step-by-step explanation: