Question

How do you differentiate `y =sqrt( x + sqrtx )` using the chain rule?

Answer 1

I get `1/(2sqrt(x+sqrt(x)))+1/(4sqrt(x^2+xsqrt(x)))`.

Explanation:

We are given: `y=sqrt(x+sqrt(x))`

The chain rule states that

`(df)/dx=(df)/(du)*(du)/dx`

So, we let `u=x+sqrt(x),:.f=sqrt(u)`.

Now, we differentiate both.

`:.(du)/dx=1+1/(2sqrt(x)`

`(df)/(du)=1/(2sqrt(u))`

So, `(df)/dx=1/(2sqrt(u))(1+1/(2sqrt(x)))`

`=1/(2sqrt(u))+1/(4sqrt(ux))`

Reversing the substitution that `u=x+sqrt(x)`, we get

`=1/(2sqrt(x+sqrt(x)))+1/(4sqrt(x(x+sqrt(x)))`

`=1/(2sqrt(x+sqrt(x)))+1/(4sqrt(x^2+xsqrt(x)))`

Answer 2

#color(blue)((1+1/(2sqrt(x)))/(2sqrt(x+sqrt(x))) #

Explanation:

For the chain rule we have:

`dy/dx=dy/(du)*(du)/(dx)`

Re-writing `sqrt(x+sqrt(x))` as `(x+x^(1/2))^(1/2)`

Letting: `u = (x+x^(1/2))`

We now have:

`y=u^(1/2)`

Using the chain rule:

`dy/dx= 1/2(u)^(-1/2)*(1+1/2x^(-1/2))`

Substituting: `u = (x+x^(1/2))`

`dy/dx= 1/2(x+x^(1/2))^(-1/2)*(1+1/2x^(-1/2))`

Simplifying:

`dy/dx= 1/2(x+x^(1/2))^(-1/2)*(1+1/2x^(-1/2))`

#color(white)(888.)=(1+1/(2x^(1/2)))/(2(x+x^(1/2))^(1/2))=(1+1/(2sqrt(x)))/(2sqrt(x+sqrt(x)))=color(blue)((1+1/(2sqrt(x)))/(2sqrt(x+sqrt(x))) #