Question

CAN SOMEONE HELP WITH THIS QUESTION?

Answer 1

Answers:

`\text{Derivative: } \ \ (dy)/(dx) = (-260x^(9) - 156x^(25)y)/(6x^(26)+7y^6)\\\\ \text{Tangent line at (1,1) is: } \ y = -32x + 33\\\\`

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Work Shown:

Let's determine the derivative dy/dx.

Part 1

`26x^(10) + 6x^(26)y+y^7 = 33\\\\ (d)/(dx)(26x^(10) + 6x^(26)y+y^7) = (d)/(dx)(33)\\\\ (d)/(dx)(26x^(10)) + (d)/(dx)(6x^(26)y)+(d)/(dx)(y^7) = 0\\\\ 10*26x^(10-1) + (d)/(dx)(6x^(26))y+(6x^(26))*(dy)/(dx)+7y^6(dy)/(dx) = 0\\\\`

Part 2

`260x^(9) + 26*6x^(26-1)y+6x^(26)*(dy)/(dx)+7y^6(dy)/(dx) = 0\\\\ 260x^(9) + 156x^(25)y+6x^(26)*(dy)/(dx)+7y^6(dy)/(dx) = 0\\\\ 260x^(9) + 156x^(25)y+(6x^(26)+7y^6)(dy)/(dx) = 0\\\\ (6x^(26)+7y^6)(dy)/(dx) = -260x^(9) - 156x^(25)y\\\\ (dy)/(dx) = (-260x^(9) - 156x^(25)y)/(6x^(26)+7y^6)\\\\`

There are many other possible ways to express the dy/dx expression.

GeoGebra and WolframAlpha are two useful tools to help verify the answer. Make sure you use the CAS mode in GeoGebra.

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Part 3

Now that we know dy/dx, we can determine the slope of the tangent at any point (x,y) on the implicit function curve.

Plug in x = 1 and y = 1.

`(dy)/(dx) = (-260x^(9) - 156x^(25)y)/(6x^(26)+7y^6)\\\\ (dy)/(dx) = (-260(1)^(9) - 156(1)^(25)(1))/(6(1)^(26)+7(1)^6)\\\\ (dy)/(dx) = (-260(1) - 156(1)(1))/(6(1)+7(1))\\\\ (dy)/(dx) = (-260 - 156)/(6+7)\\\\ (dy)/(dx) = (-416)/(13)\\\\ (dy)/(dx) = -32\\\\`

The slope of the tangent line at (1,1) is m = -32.

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Part 4

Apply the point-slope formula to determine the tangent line.

`m = -32 = \text{ slope}\\(x_1,y_1) = (1,1) = \text{the point the tangent line goes through}`

So,

`y - y_1 = m(x - x_1)\\\\y - 1 = -32(x - 1)\\\\y - 1 = -32x + 32\\\\y = -32x + 32 + 1\\\\y = -32x + 33\\\\`