Use implicit differentiation to find the slope of the tangent line to the curve
y/x+6y=x^2–6 at the point (1,–5/31) .
Again i think i'm messing up with the algebra here. I used quotient rule to get
[(x+6y)(y')-(y)(1+6y')]/(x+6y)^2=2x
I don't know how to go from here to find the m. Please help :)
Your derivative equation looks good.
sub in x=1 and y = -5/31 into that derivative equation, then solve it for y'
that will be your m
Good luck with that messy arithmetic.
I got y'=(2x+x^2+12yx+6y^2-y)/(x+6y+6)
But when i plugged in the numbers it came out right... do you think the equation is wrong or i just typed in the numbers wrong?
BTW thanks for replying!
using your [(x+6y)(y')-(y)(1+6y')]/(x+6y)^2=2x
we can continue
xy' + 6yy' - y - 6yy' = 2x
xy' - y = 2x
y' = (2x+y)/x
using the given point:
y' = (2 - 5/31)/1 = 57/31
y + 5/31 = (57/31)(x-1)
31y + 5 = 57x - 57
57x - 31y = 62
check my arithmetic.
I did this exactly with this question and i got the wrong answer.
Use implicit differentiation to find the slope of the tangent line to the curve
y/(x+5y)=x^3+5
at the point (x=1, y=6/–29).
But I noticed after "we can continue..."
the denominator was moved to the right side of the equation.. so maybe that's it..??
Sorry. I meant the denominator from the quotient rule was never multiplied to the right or so. I don't see it. Where did that portion go?
To find the slope of the tangent line to the curve, we can use implicit differentiation. Let's go step by step to solve it together:
Step 1: Start with the given equation: y/x + 6y = x^2 - 6.
Step 2: Differentiate both sides of the equation with respect to x. Treat y as a function of x and use the chain rule for terms involving y.
For the left side (y/x), we can use the quotient rule, which states:
(differentiation of numerator * denominator - numerator * differentiation of denominator) / (denominator^2)
Applying the quotient rule to y/x, we get:
[(x * (d/dx) y - y * (d/dx) x) / (x^2)] + 6 * (d/dx) y = (d/dx) (x^2 - 6)
Step 3: Simplify the equation using the derivative rules and algebra.
For the left side, we have:
[(x * (d/dx) y - y * (d/dx) x) / (x^2)] + 6 * (d/dx) y
(d/dx) x is simply 1, and (d/dx) y is denoted as y'.
Rewriting the equation, we have:
[(x * y') - y] / x^2 + 6y' = 2x
Step 4: Rearrange terms to isolate y' (the derivative of y with respect to x), which represents the slope of the tangent line.
To do that, move the terms containing y' to one side and the remaining terms to the other side:
[(x * y') + 6y'] = 2x - [(y) / x^2]
Combine like terms:
(x * y' + 6y') = 2x - (y / x^2)
Factor out y' from the left side:
y' (x + 6) = 2x - (y / x^2)
Since we want the slope of the tangent line at a specific point (1, -5/31), we substitute x = 1 and y = -5/31 into the equation:
(-5/31)' (1 + 6) = 2(1) - (-5/31) / (1^2)
-5/31 * 7y' = 2 + 5/31
Step 5: Solve for y' to find the slope of the tangent line. Rearrange the equation:
-5/31 * 7y' = 2 + 5/31
Multiply both sides by -1/7:
-5/31 * 7y' * (-1/7) = (2 + 5/31) * (-1/7)
Simplify:
y' = - (2 + 5/31) / (5/31)
Using the arithmetic operations:
y' = - (62 + 5) / 31 = -67 / 31
Therefore, the slope of the tangent line to the curve at the point (1, -5/31) is -67/31.