# Stewart – Calculus – 3.5 – Implicit Differentiation and Derivatives of Inverse Trigonometric Functions

Find $y''$ by implicit differentiation:
$9x^{2}+y^{2}=9$.

$18x+2yy'=0$
$2yy'=-18x$
$y'=-9x/y$

$y''=-9(\frac{y\cdot 1\text{--}x\cdot y'}{y^{2}})$

$y''=-9(\frac{y\cdot 1\text{--}x(-9x/y)}{y^{2}})$ ($y'$ is replaced with $-9x/y$)

$y''=-9(\frac{y^{2}+9x^{2}}{y^{3}})$ (after multiplying by $y$ to eliminate denominator from $-9x$)

$y''=-9(\frac{9}{y^{3}})$ (notice that $y^{2}+9x^{2}$ equals the original equation)

So, $y''=\frac{-81}{y^{3}}$.

Derivatives of Inverse Trigonometric Functions