- #1

chwala

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- Homework Statement
- See attached question and ms

- Relevant Equations
- differentiation

The question and ms guide is pretty much clear to me. I am attempting to use a non-implicit approach.

##\tan y=x, ⇒y=\tan^{-1} x##

We know that ##1+ \tan^2 x= \sec^2 x##

##\dfrac{dx}{dy}=sec^2 y##

##\dfrac{dx}{dy}=1+\tan^2 y##

##\dfrac{dy}{dx}=\dfrac{1}{1+x^2}##

##\dfrac{d^2y}{dx^2}=\dfrac{(1+x)0-1(2x)}{(1+x^2)^2}##

##\dfrac{d^2y}{dx^2}=\dfrac{-2x}{(1+x^2)^2}##

##\dfrac{d^2y}{dx^2}=-2x ⋅\left(\dfrac{dx}{dy}\right)^2##

No question here ...just looking at the problem from a different perspective ...any insight is welcome...

We know that ##1+ \tan^2 x= \sec^2 x##

##\dfrac{dx}{dy}=sec^2 y##

##\dfrac{dx}{dy}=1+\tan^2 y##

##\dfrac{dy}{dx}=\dfrac{1}{1+x^2}##

##\dfrac{d^2y}{dx^2}=\dfrac{(1+x)0-1(2x)}{(1+x^2)^2}##

##\dfrac{d^2y}{dx^2}=\dfrac{-2x}{(1+x^2)^2}##

##\dfrac{d^2y}{dx^2}=-2x ⋅\left(\dfrac{dx}{dy}\right)^2##

No question here ...just looking at the problem from a different perspective ...any insight is welcome...