Tangent line to the curve.

**Zanderist** · March 16th 2010, 06:10 PM

How do I do the following.

find the equation of the tangent line to the curve:

$xy^3+xy=14$

At point (7,1).

**pickslides** · March 16th 2010, 06:16 PM

Originally Posted by Zanderist

How do I do the following.

find the equation of the tangent line to the curve:

$xy^3+xy=14$

At point (7,1).

$x(y^3+y)=14$

$x=\frac{14}{y^3+y}$

Now find $\frac{dx}{dy}$ at $y = 1$

**skeeter** · March 16th 2010, 06:16 PM

Originally Posted by Zanderist

How do I do the following.

find the equation of the tangent line to the curve:

$xy^3+xy=14$

At point (7,1).

find $\frac{dy}{dx}$ in terms of $x$ and $y$ , then sustitute $7$ for $x$ and $1$ for $y$ in the derivative expression to find the slope at that point.

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