Use implicit differentiation to find the slope of the tangent line to the curve

Question

Use implicit differentiation to find the slope of the tangent line to the curve

y/(x-9y) = x^8 -3

at the point (1, 2/17)

find the slope.

Answer 1

y = (x⁸ - 3)(x - 9y); y' = 8x⁷(x - 9y) + (x⁸ -3)(1 - 9y'); At point (1, 2/17) we have: y' = 8(1 - 18/17) - 2( 1 - 9y');

-17y' = - 42/17; y' = 42/289

Answer 2

To find the slope of the tangent line using implicit differentiation, start with the given equation:

\[

\frac{y}{x - 9y} = x^8 - 3

\]

You need to differentiate both sides with respect to \(x\). Since \(y\) is a function of \(x\), you need to apply the chain rule when differentiating any terms involving \(y\).

Step 1: Differentiate the left side

\[

\text{Left side: } \frac{y}{x - 9y}

\]

You need to use the quotient rule for differentiation, which is:

\[

\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

\]

Here, \(u = y\) and \(v = x - 9y\), so:

\[

\frac{du}{dx} = \frac{dy}{dx} \quad \text{and} \quad \frac{dv}{dx} = 1 - 9\frac{dy}{dx}

\]

Using the quotient rule:

\[

\frac{d}{dx} \left(\frac{y}{x - 9y}\right) = \frac{(x - 9y) \cdot \frac{dy}{dx} - y \cdot (1 - 9\frac{dy}{dx})}{(x - 9y)^2}

\]

Simplifying:

\[

= \frac{(x - 9y)\frac{dy}{dx} - y + 9y\frac{dy}{dx}}{(x - 9y)^2}

\]

\[

= \frac{(x - 9y + 9y)\frac{dy}{dx} - y}{(x - 9y)^2}

\]

\[

= \frac{x\frac{dy}{dx} - y}{(x - 9y)^2}

\]

Step 2: Differentiate the right side

\[

\text{Right side: } x^8 - 3

\]

The derivative of \(x^8\) with respect to \(x\) is \(8x^7\), and the derivative of \(-3\) is 0, so:

\[

\frac{d}{dx}(x^8 - 3) = 8x^7

\]

Step 3: Equate the derivatives

Now, equate the derivatives from both sides:

\[

\frac{x\frac{dy}{dx} - y}{(x - 9y)^2} = 8x^7

\]

Step 4: Solve for \(\frac{dy}{dx}\)

To solve for \(\frac{dy}{dx}\), multiply both sides by \((x - 9y)^2\) to clear the denominator:

\[

x\frac{dy}{dx} - y = 8x^7(x - 9y)^2

\]

Isolate \(\frac{dy}{dx}\):

\[

x\frac{dy}{dx} = y + 8x^7(x - 9y)^2

\]

\[

\frac{dy}{dx} = \frac{y + 8x^7(x - 9y)^2}{x}

\]

Step 5: Substitute the given point \((x, y) = \left(1, \frac{2}{17}\right)\)

Substitute \(x = 1\) and \(y = \frac{2}{17}\) into the equation:

\[

\frac{dy}{dx} = \frac{\frac{2}{17} + 8(1)^7(1 - 9\left(\frac{2}{17}\right))^2}{1}

\]

Simplify the expression inside:

\[

1 - 9\left(\frac{2}{17}\right) = 1 - \frac{18}{17} = \frac{17}{17} - \frac{18}{17} = -\frac{1}{17}

\]

Square this result:

\[

\left(-\frac{1}{17}\right)^2 = \frac{1}{289}

\]

Now:

\[

\frac{dy}{dx} = \frac{\frac{2}{17} + 8 \cdot \frac{1}{289}}{1}

\]

\[

\frac{dy}{dx} = \frac{\frac{2}{17} + \frac{8}{289}}{1}

\]

To combine the fractions, get a common denominator:

\[

\frac{2}{17} = \frac{34}{289}, \quad \text{so}

\]

\[

\frac{dy}{dx} = \frac{\frac{34}{289} + \frac{8}{289}}{1} = \frac{\frac{42}{289}}{1}

\]

Thus, the slope of the tangent line at the point \((1, \frac{2}{17})\) is:

\[

\boxed{\frac{42}{289}}

\]

Please note that the above text is given in LaTeX format. Please convert it into Word or PDF format to see equations more clearly.

Answer 3

y / (x-9y) = x⁸-3

[(x-9y)(dy/dx) - y(1-9(dy/dx)] / (x-9y)² = 8x⁷

Plug in the given point: [(-1/17)(dy/dx) - (2/17)(1-9(dy/dx))] / (-1/17)² = 8

-17(dy/dx) - 34(1-9(dy/dx)) = 8

289(dy/dx) = 42

dy/dx = 42/289