How do you show that if #y=sqrt(x+sqrt(x^2+1)#, then #y'=(sqrt(x+sqrt(x^2+1))/(2sqrt(x^2+1)))# ?

1 Answer
Sean
Feb 4, 2018

Please see below.

Explanation:

.

#y=sqrt(x+sqrt(x^2+1))#

#y=(x+sqrt(x^2+1))^(1/2)#

Let's let #u=x+sqrt(x^2+1)=x+(x^2+1)^(1/2)#

And let's let #z=x^2+1#

#dz=2xdx#

Now, we substitute in #u#

#u=x+z^(1/2)#

Take the derivative of both sides:

#du=dx+1/2z^(-1/2)dz#

Substitute back for #z# and #dz#:

#du=dx+1/2(x^2+1)^(-1/2)(2xdx)#

Divide both sides by #dx#:

#(du)/dx=1+x/sqrt(x^2+1)#

#y=u^(1/2)#

#dy/(du)=1/2u^(-1/2)#

The Chain Rule says:

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

Let's plug them in:

#dy/dx=1/2u^(-1/2)(1+x/sqrt(x^2+1))#

Let's substitute back for #u#:

#dy/dx=1/2(x+sqrt(x^2+1))^(-1/2)(1+x/sqrt(x^2+1))#

#dy/dx=(1/(2sqrt(x+sqrt(x^2+1))))((sqrt(x^2+1)+x)/sqrt(x^2+1))#

Now, if we multiply the right had side by:

#(sqrt(sqrt(x^2+1)-x))/(sqrt(sqrt(x^2+1)-x))#

we get:

#dy/dx=((sqrt(sqrt(x^2+1)-x))(sqrt(x^2+1)+x))/((sqrt((x^2+1)-x^2))(2sqrt(x^2+1))#

#dy/dx=((sqrt(sqrt(x^2+1)-x))sqrt((sqrt(x^2+1)+x)^2))/(2sqrt(x^2+1))#

#dy/dx=((sqrt(sqrt(x^2+1)-x))sqrt(2x^2+2xsqrt(x^2+1)+1))/(2sqrt(x^2+1))#

#dy/dx=sqrt(2x^2sqrt(x^2+1)+2x^3+2x+sqrt(x^2+1)-2x^3-2x^2sqrt(x^2+1)-x)/(2sqrt(x^2+1)#

#dy/dx=sqrt(cancelcolor(red)(2x^2sqrt(x^2+1))cancelcolor(red)(+2x^3)+2x+sqrt(x^2+1)cancelcolor(red)(-2x^3)cancelcolor(red)(-2x^2sqrt(x^2+1))-x)/(2sqrt(x^2+1)#

#dy/dx=sqrt(x+sqrt(x^2+1))/(2sqrt(x^2+1)#

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