Implicit differentiation

**juventinoalex** · October 5th 2010, 07:11 PM

x^3y^4-9xsiny=14
I took the derivitives as
(3x^2y^4)+((x^3)4y^3(y`))-(9siny+9xcosy(y`))
I solved for y` and got
y` =(3x^2y^4+9siny)/x^3(4y^3)-9cosy)
But this doesn't come out as correct.

Any suggestions?
Thanks

**Soroban** · October 5th 2010, 08:42 PM

Hello, juventinoalex!

$x^3y^4 - 9x\sin y \:=\:14$

$\text{I differentiated: }\;3x^2y^4 + x^34y^3(y')-\bigg(9\sin y+9x\cos y(y')\bigg) \:=\:0$ . Yes

$\text{I solved for }y'\text{ and got: }\;y' \:=\:\dfrac{3x^2y^4 + 9\sin y}{4x^3y^3-9\cos y}$ . No

$\text{But this doesn't come out as correct.}$

We have: . $3x^2y^4 - 4x^3y^3(y') - 9\sin y - 9x\cos y (y') \;=\;0$

. . .Then: . $4x^3y^3(y') + 9x\cos y(y') \;=\;3x^2y^4 - 9\sin y$

. . Factor: . $(4x^3y^3 + 9x\cos y)(y') \;=\;3x^2y^4 - 9\sin y$

Therefore: . $y' \;=\;\dfrac{3x^2y^4 - 9\sin y}{4x^3y^3 + 9x\cos y}$

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