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

httpv://www.youtube.com/watch?v=5yTVUZCaU6k

httpv://www.youtube.com/watch?v=2dv_PfEFZXY

httpv://www.youtube.com/watch?v=ximF06lmqPM

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

httpv://www.youtube.com/watch?v=xBI8hVBB0pQ

Derivatives of Inverse Trigonometric Functions

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