implicit differentiation...02

**Foxlion** · March 23rd 2011, 02:46 PM

problem: $1+x = sin(xy^2)$

solution attempt: differentiating both sides, I have
$1 = -cosx \cdot\frac{d}{dx}[xy^2]$

$1 = -cosx(y^2+2xyy')$

dividing by $-cosx$ on both sides, I have
$-\frac{1}{cosx} =y^2+2xyy'$

dividing by $2xy$ on both sides, I have
$-\frac{2xy}{cosx} = \frac{y^2}{2xy} + y'$

subtracting $\frac{y^2}{2xy}$ from both sides, I have
$y' = -\frac{2xy}{cosx}-\frac{y^2}{2xy}$

then,
$y' = -\frac{y^2(4x^2 - cosx)}{2xycosx}$

**adkinsjr** · March 23rd 2011, 03:01 PM

You differentiated the right side wrong.

$\frac{d}{dx}sin(xy^2)=cos(xy^2)\frac{d}{dx}(xy^2)= (y^2+2xyy')cos(xy^2)$

Remember when you use the chain rule you have to keep whatever is in the parenthesis.

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