What is the equation of the line normal to  f(x)=sqrt(1/(x^2+2)  at  x=1?

Jun 2, 2016

$y - \sqrt{\frac{1}{3}} = \sqrt[3]{9} \left(x - 1\right)$

Explanation:

in simplifying the equation we get

${\left({x}^{2} + 2\right)}^{- \frac{1}{2}}$

We then apply the chain rule to get

-x/(root(3)(x^2+2)^2

Then we substitute in 1 for x to find the gradient of the tangent at x=1

This results us getting $- \frac{1}{\sqrt[3]{9}}$, however we need the normal's gradient, not the tangent's. The normal times the tangent gives us -1, so therefore the gradient of the normal will be $\sqrt[3]{9}$.

Now we know the gradient, we just need the co-ordinates which it runs through. We then sub x=1 back int the original equation to find the point which the normal will pass through. The y value, after subsitution is $\sqrt{\frac{1}{3}}$.

Therefore by substituting the numbers/values into the point gradient formula, we'll find the equation of the normal. The point gradient formula is

$y - {y}_{1} = m \left(x - {x}_{1}\right)$