How do you find the derivative of #sqrt(x^2+1)#?

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Answer 1 · 2016-11-01T01:43:08

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

#y^2=x^2+1#

Use implicit differentiation on the left hand side of the equation and ordinary differentiation on the right hand side of the equation.

#2y*(dy)/(dx)=2x#

#y*(dy)/(dx)=x#

#(dy)/(dx)=x/y#

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

Answer 2 · 2016-11-01T01:53:20

The power rule says that #d/dx(x^n) = nx^(n-1)#

The chain rule, when combined with the power rule (sometimes called "the general power rule" says that

#d/dx(u^n) = n u^(n-1) (du)/dx#

So

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

# = 1/2(x^2+1)^(1/2-1) * d/dx(x^2+1)#

# = 1/2 (x^2+1)^(-1/2)* (2x)#

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

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

The differentiation is sped up considerably by learning the #d/dx(sqrtx) = 1/(2sqrtx)#.

After this is learned, we can simply write

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