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

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

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