Question on implicit differentiation

I'm new to the forum and having a tough time on some of my homework problems and was looking for some feedback.

So here's one of my problems.

Use implicit differentiation to find the slope of the tangent line to the curve

$\displaystyle \frac{y}{x-6y}=x^3-3$

at the point $\displaystyle (1,\frac{-2}{-11})$

I'm just having trouble finding a starting point. Any help is greatly appreciated.

Quote:

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Originally Posted by ascendancy523

I'm new to the forum and having a tough time on some of my homework problems and was looking for some feedback.

So here's one of my problems.

Use implicit differentiation to find the slope of the tangent line to the curve

$\displaystyle \frac{y}{x-6y}=x^3-3$

at the point $\displaystyle (1,\frac{-2}{-11})$

I'm just having trouble finding a starting point. Any help is greatly appreciated.

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$\displaystyle y =(x^3-3)(x-6y)$

$\displaystyle y = x^4 - 6x^3y - 3x + 18y$

$\displaystyle x^4 - 6x^3y - 3x + 17y = 0$

now take the derivative w/r to x ... should be rather straightforward. remember to use the product rule for the second term.

once you determine an expression for $\displaystyle \frac{dy}{dx}$, sub in your point's coordinate values to find the slope