Asked by Sarah

If ln(x^2-15y)=x-5+5 and y(-4)=1 find Y prime(-4) by implicit differentiation

thus the equation of the tangent line to the graph at the point (-4,1) is y=

Answers

Answered by bobpursley
What is this x-5+5?
Answered by Steve
Hmmm. Let me take a stab at it. AT (-4,1) we have

ln(x^2-15y) = ln(1) = 0

Hmmph. Can't make out a reasonable interpretation that is zero on the right.

Anyway, regardless, let's just say that it is ok as it is. Then we have, taking derivatives of both sides, using the chain rule as needed:

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

So, fix the right side as needed, and solve for y', then plug in (-4,1) to get the numeric value.
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