Use implicit differentiation to find the slope of the tangent line to the curve
y/(x–9y)=x^6–2
at the point (1,-1/-8).
Can someone please help me?
I don't understand
puzzled, why would your write the point as
(1 , -1/-8) and not as simply (1 , 1/8) ?
first expand it
y = x^7 - 2x - 9x^6 y + 18y
0 = x^7 - 2x - 9x^6 y + 17y
0 = 7x^6 - 2 - 9x^6 dy/dx - 54x^5y + 17dy/dx
dy/dx(17 - 9x^6) = 54x^5y - 7x^6
dy/dx = (54x^5y - 7x^6))/(17 - 9x^6)
at x = 1
dy/dx = (54(1/8) -7)/(17 - 9)
= (-1/4) / 8
= -1/32
equation of tangent:
y + 1/8 = (-1/32)(x-1)
y = (-1/32)x + 1/32 - 1/8
y = (-1/32)x -3/32
check my arithmetic
To find the slope of the tangent line to the curve, we can use the method of implicit differentiation.
Given the equation: y/(x–9y) = x^6–2
To differentiate implicitly, we treat y as a function of x and differentiate both sides of the equation with respect to x.
We start by differentiating the left-hand side of the equation. Since we have y appearing in the numerator and denominator, we will apply the quotient rule.
The quotient rule states that for a function u(x) divided by a function v(x), the derivative is given by:
(d/dx)(u/v) = (v * du/dx - u * dv/dx) / v^2
Let's apply the quotient rule to the left-hand side of the equation:
(d/dx)(y/(x–9y)) = [(x–9y) * d/dx(y) - y * d/dx(x–9y)] / (x–9y)^2
Now, we need to find the derivatives of y with respect to x and x–9y with respect to x.
The derivative of y with respect to x is denoted as dy/dx or y'.
The derivative of x–9y with respect to x can be found using the chain rule, which states that for a composite function f(g(x)), the derivative is given by f'(g(x)) * g'(x).
Let's find the derivatives:
dy/dx = y' (derivative of y with respect to x)
d/dx (x–9y) = d/dx (x) - 9 * d/dx (y)
= 1 - 9 * dy/dx
Now, let's substitute these derivatives back into the equation:
[(x–9y) * dy/dx - y * (1 - 9 * dy/dx)] / (x–9y)^2 = x^6–2
Next, we substitute the given point (1, -1/8) into the equation to find the value of dy/dx at that point.
[(1 – 9 * (-1/8)) * dy/dx - (-1/8) * (1 - 9 * dy/dx)] / (1 – 9 * (-1/8))^2 = 1^6–2
Simplifying the equation and substituting the given values, we can solve for dy/dx:
[(1 + 9/8) * dy/dx + (1/8) * (1 + 9 * dy/dx)] / (1 + 9/8)^2 = 1
[(17/8) * dy/dx + (1/8) * (1 + 9 * dy/dx)] / (17/8)^2 = 1
[(17/8) * dy/dx + (1/8) * (1 + 9 * dy/dx)] / (289/64) = 1
Multiplying both sides of the equation by (289/64) gives:
(17/8) * dy/dx + (1/8) * (1 + 9 * dy/dx) = 289/64
Multiplying through the parentheses gives:
17/8 * dy/dx + 1/8 + 9/8 * dy/dx = 289/64
Combining like terms gives:
26/8 * dy/dx + 1/8 = 289/64
26/8 * dy/dx = 289/64 - 1/8
26/8 * dy/dx = (289 - 8) / 64
26/8 * dy/dx = 281/64
To solve for dy/dx, we multiply both sides by 8/26:
dy/dx = (281/64) * (8/26)
dy/dx = 281/208
Therefore, the slope of the tangent line to the curve at the point (1, -1/8) is 281/208.
Of course! I'll walk you through the process.
To find the slope of the tangent line to the curve at the given point using implicit differentiation, follow these steps:
Step 1: Differentiate both sides of the equation with respect to x.
The given equation is:
y/(x – 9y) = x^6 – 2
To differentiate the left side of the equation, you can use the quotient rule, which states that for any two functions u(x) and v(x), the derivative of their quotient is (v * u' - u * v') / (v^2). Let's apply this rule:
(differentiate numerator) * (denominator) - (numerator) * (differentiate denominator)
_____________________________________________________________
(denominator)^2
Differentiating the left side using the quotient rule:
(y)' * (x – 9y) – y * (x – 9y)'
Step 2: Implicitly differentiate the right side of the equation.
Differentiate each term on the right side of the equation with respect to x:
d/dx (x^6) = 6x^5
d/dx (-2) = 0 (since a constant is always zero when differentiated)
Step 3: Simplify the differentiated equation.
Combining the results from Steps 1 and 2:
(y)' * (x – 9y) – y * (x – 9y)' = 6x^5
We now have a differential equation:
(y') * (x – 9y) – y * (x – 9y)' = 6x^5
Step 4: Substitute the given coordinates into the equation.
The point given is (1, -1/-8), which can be simplified to (1, 8).
Substituting x = 1 and y = 8 into the equation:
(y') * (1 – 9(8)) – 8 * (1 – 9(8))' = 6(1^5)
Simplifying further:
(y') * (1 – 72) – 8 * (1 – 72)' = 6
Step 5: Solve for (y') to find the slope of the tangent line.
Simplifying the equation:
(y') * (-71) – 8 * (-71)' = 6
(y') * (-71) + 8 * 1 = 6
(y') * (-71) + 8 = 6
(y') * (-71) = -2
Now, divide both sides by -71:
y' = -2 / (-71)
y' = 2 / 71
Therefore, the slope of the tangent line to the curve at the point (1, 8) is 2 / 71.