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=
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)
Well, well, well, it looks like we have a math question here. But fear not, my friend, for Clown Bot is here to add a touch of humor to your calculus woes.
Let's start by differentiating the given equation implicitly with respect to x:
d/dx[ln(x^2 - 15y)] = d/dx[x - y + 5]
Now, I could go on and bore you with all the steps and calculations, but where's the fun in that? Instead, I'll simply give you the result of implicit differentiation:
(2x - 15y)/(x^2 - 15y) = 1 - dy/dx
Okay, now let's find the value of dy/dx when y = -4:
(2x - 15(-4))/(x^2 - 15(-4)) = 1 - dy/dx
(2x + 60)/(x^2 + 60) = 1 - dy/dx
Since the point (-4, 1) lies on the graph, we can substitute these values into the equation to find dy/dx:
(2(-4) + 60)/((-4)^2 + 60) = 1 - dy/dx
(-8 + 60)/(16 + 60) = 1 - dy/dx
52/76 = 1 - dy/dx
dy/dx = 1 - 52/76
dy/dx = 76/76 - 52/76
dy/dx = 24/76
Now that we have dy/dx, we can find the equation of the tangent line using the point-slope form:
y - y1 = m(x - x1)
Plugging in the values (-4, 1) and dy/dx = 24/76, we get:
y - 1 = (24/76)(x - (-4))
y - 1 = (24/76)(x + 4)
And finally, after a little bit of simplification, we arrive at:
y = (6/19)(x + 4) + 1
Voila! The equation of the tangent line to the graph at point (-4, 1) is y = (6/19)(x + 4) + 1.
To find the equation of the tangent line to the graph at the point (-4,1), we need to use implicit differentiation.
Step 1: Differentiate both sides of the equation ln(x^2-15y)=x-y+5 with respect to x.
d/dx[ln(x^2-15y)] = d/dx[x-y+5]
Step 2: Apply the chain rule on the left-hand side and differentiate each term on the right-hand side.
(1/(x^2-15y))*d/dx[(x^2-15y)] = 1 - 1
Step 3: Simplify the equation.
(1/(x^2-15y))*(2x - 15(dy/dx)) = 0
Step 4: Solve for dy/dx.
2x - 15(dy/dx) = 0
dy/dx = 2x/15
Step 5: Evaluate dy/dx at the point (-4,1).
dy/dx = 2(-4)/15
dy/dx = -8/15
Step 6: Use the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point.
y - 1 = (-8/15)(x - (-4))
Step 7: Simplify the equation.
y - 1 = (-8/15)(x + 4)
Step 8: Convert the equation to slope-intercept form.
y = -8/15(x + 4) + 1
Step 9: Simplify further to obtain the equation of the tangent line.
y = -8/15x - 32/15 + 1
y = -8/15x - 17/15
Therefore, the equation of the tangent line to the graph at point (-4,1) is y = -8/15x - 17/15.
To find the equation of the tangent line to the graph at the point (-4, 1), we'll need to follow these steps:
1. Differentiate both sides of the equation with respect to x using implicit differentiation.
2. Solve for dy/dx to find the slope of the tangent line.
3. Plug in the given point (-4, 1) and the slope into the equation y-y1 = m(x-x1) to find the equation of the tangent line.
Let's go through each step:
1. Differentiate both sides of the equation ln(x^2 - 15y) = x - y + 5 with respect to x using implicit differentiation.
First, let's differentiate ln(x^2 - 15y) with respect to x:
d/dx[ln(x^2 - 15y)] = d/dx(x - y + 5)
Using the chain rule, the left side becomes:
1 / (x^2 - 15y) * d/dx(x^2 - 15y) = 1 - dy/dx
Differentiating the right side with respect to x gives us:
1 - dy/dx
2. Solve for dy/dx to find the slope of the tangent line.
Now, let's focus on isolating dy/dx. Rearrange the equation to isolate dy/dx:
1 - dy/dx = 1 / (x^2 - 15y) * d/dx(x^2 - 15y)
Multiply both sides by (x^2 - 15y):
(x^2 - 15y) - dy/dx(x^2 - 15y) = d/dx(x^2 - 15y)
Apply the product rule to differentiate x^2 - 15y:
(x^2 - 15y) - dy/dx(x^2 - 15y) = 2x - 15(dy/dx)
Move the dy/dx terms to one side of the equation:
dy/dx(x^2 - 15y) - 15(dy/dx) = 2x - (x^2 - 15y)
Factor out dy/dx:
dy/dx[(x^2 - 15y) - 15] = 2x - (x^2 - 15y)
Combine like terms:
dy/dx(x^2 - 15y - 15) = 2x - x^2 + 15y
Divide both sides by (x^2 - 15y - 15) to solve for dy/dx:
dy/dx = (2x - x^2 + 15y) / (x^2 - 15y - 15)
3. Plug in the given point (-4, 1) and the slope into the equation y - y1 = m(x - x1) to find the equation of the tangent line.
Using the point (-4, 1), substitute x1 = -4 and y1 = 1 into the equation:
y - 1 = dy/dx(-4)(x + 4)
Substitute the slope dy/dx we found earlier:
y - 1 = [(2x - x^2 + 15y) / (x^2 - 15y - 15)](-4)(x + 4)
Now, simplify and solve for y to get the equation of the tangent line.