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=
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.
To find y'(-4) by implicit differentiation, we will differentiate both sides of the equation ln(x^2-15y) = x-5+5 with respect to x using the chain rule.
First, let's differentiate the left side of the equation:
d/dx[ln(x^2-15y)] = d/dx[x-5+5]
We have to use the chain rule because we have the natural logarithm function. The derivative of ln(u) is 1/u multiplied by the derivative of u with respect to x.
So, applying the chain rule to the left side of the equation:
(1/(x^2-15y)) * d/dx[(x^2-15y)] = 1
Now, let's differentiate the right side of the equation:
d/dx[x-5+5] = d/dx[x]
This is simply the derivative of x with respect to x, which is 1.
Now, simplifying the equation:
(1/(x^2-15y)) * (2x - 15 * (dy/dx)) = 1
Multiplying through by (x^2-15y) to eliminate the fraction:
2x - 15 * (dy/dx) = x^2-15y
To find the derivative dy/dx, we isolate it by rearranging the equation:
15 * (dy/dx) = x^2 - 2x - 15y
Now, we can substitute x = -4 and y = 1 into this equation to find dy/dx at (-4,1):
15 * (dy/dx) = (-4)^2 - 2(-4) - 15(1)
15 * (dy/dx) = 16 + 8 - 15
15 * (dy/dx) = 9
(dy/dx) = 9/15 = 3/5
So, y'(-4) = 3/5.
To find the equation of the tangent line to the graph at the point (-4,1), we can use the point-slope formula.
The equation of a line can be written as: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Using the point (-4,1) and the slope m = 3/5, we have:
y - 1 = (3/5)(x - (-4))
Simplifying further:
y - 1 = (3/5)(x + 4)
Expanding:
y - 1 = (3/5)x + 12/5
Moving the constant to the other side:
y = (3/5)x + 12/5 + 1
y = (3/5)x + 12/5 + 5/5
y = (3/5)x + 17/5
Therefore, the equation of the tangent line to the graph at the point (-4,1) is y = (3/5)x + 17/5.
To find y'(-4) by implicit differentiation, you need to differentiate both sides of the given equation with respect to x and then solve for y'. Let's go step by step:
1. Start with the given equation:
ln(x^2-15y) = x-5+5
2. Differentiate both sides of the equation with respect to x using the chain rule:
1/(x^2-15y) * (2x - 15y') = 1
3. Simplify the equation:
2x - 15y' = x^2 - 15y
4. Rearrange the equation to solve for y':
15y' = x^2 - 2x - 15y
y' = (x^2 - 2x - 15y) / 15
5. Now, we need to evaluate y' at x = -4 and y = 1. Substitute these values into the equation:
y'(-4) = (-4^2 - 2 * -4 - 15 * 1) / 15
y'(-4) = (16 + 8 - 15) / 15
y'(-4) = 9/15
y'(-4) = 3/5
So, y'(-4) is equal to 3/5.
To find the equation of the tangent line to the graph at the point (-4, 1), we have the slope of the tangent line (y'(-4) = 3/5) and the point (-4, 1).
The point-slope form of a linear equation is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Plugging in the values, we get:
y - 1 = (3/5)(x - (-4))
Simplifying the equation:
y - 1 = (3/5)(x + 4)
5y - 5 = 3(x + 4)
5y - 5 = 3x + 12
3x - 5y = -17
Therefore, the equation of the tangent line to the graph at the point (-4, 1) is 3x - 5y = -17.