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=
2 answers
What is this x-5+5?
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.
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.