Are these derivatives correct?

**Algebrah** · October 21st 2011, 08:07 AM

Find the derivative of $(xy+1)^3 = x - y^2 + 8$ using implicit differentiation. Do not simplify.

$(xy+1)^3 = x - y^2 + 8$

$3(xy+1)^2 \cdot (y + x \cdot \frac{dy}{dx}) = 1 - 2y \cdot \frac{dy}{dx}$

$y + x \cdot \frac{dy}{dx} + \frac{2y \cdot \frac{dy}{dx}}{3(xy+1)^2} = \frac {1}{3(xy+1)^2}$

$x \cdot \frac{dy}{dx} + \frac{2y \cdot \frac{dy}{dx}}{3(xy+1)^2} = \frac {1}{3(xy+1)^2} -y$

$\frac{dy}{dx}[x + \frac{2y}{3(xy+1)^2}] = \frac {1}{3(xy+1)^2} -y$

$\frac{dy}{dx} = \frac {\frac {1}{3(xy+1)^2} -y}{x + \frac{2y}{3(xy+1)^2}}$

The question stated not to simplify, so this would be my final answer.

Find the derivative of $\sin(xy) + x = e^y$ using implicit differentiation. Do not simplify.

$\sin(xy) + x = e^y$

$\cos(xy) \cdot (y + x \cdot \frac{dy}{dx}) + 1 = e^y \cdot \frac{dy}{dx}$

$\cos(xy) \cdot (y + x \cdot \frac{dy}{dx}) - e^y \cdot \frac{dy}{dx} = -1$

$x \cdot \frac{dy}{dx} - \frac{e^y \cdot \frac{dy}{dx}}{\cos(xy)} = \frac{-1}{\cos(xy)} - y$

$\frac{dy}{dx}[x - \frac{e^y}{\cos(xy)}] = \frac{-1}{\cos(xy)} - y$

$\frac{dy}{dx} = \frac{\frac{-1}{\cos(xy)} - y}{[x - \frac{e^y}{\cos(xy)}]}$

I opted to not use logarithmic differentiation for some reason. Is this solution still valid?

Thanks a lot!

**SammyS** · October 21st 2011, 08:34 AM

They look fine to me.

I don't think that logarithmic differentiation would have helped .

**Algebrah** · October 21st 2011, 08:40 AM

Thanks for the response!

The other people in my class that I talked to said that they used logarithmic differention; I haven't tried it yet with this one yet.

Though, as long as they're correct, that's all that matters at the moment.

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