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


Derivatives of Inverse Trigonometric Functions

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