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

2 Answers
Mar 17, 2018

Answer:

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)))#

Mar 17, 2018

Answer:

#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))) #