# Calcalus

If ln(x^2-15y)=x-y+5 and y(-4) by implicit differentation =
Thus the equation of the tangent line to the graph at point (-4,1) is
y=

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1. ln(x^2-15y)=x-y+5
1/(x^2-15y) * (2x-15y') = 1-y'
2x-15y' = x^2-15y - (x^2-15y)y'

Now just collect terms and solve for y' to get

y' = (x^2-2x-15y)/(x^2-15y-15)

So, at (-4,1) that is

y'(-4) = (16+8-15)/(16-15-15) = -9/14

Finally, using the point-slope form for a line, you have the tangent line as

y-1 = -9/14 (x+4)

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