What is the derivative of y = (√x -2) / (3√x)? ?

The full question is as follows: Write an equation of the tangent to each curve at the given point. The point given is (1, -1). I have no issues with the rest of the question, just finding the derivative of the given equation. 

Update:

The Answer is 5x - 6y - 11 = 0

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  • 1 month ago
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    f(x) = (√x - 2)/(³√x) ← this i the function

    f(1) = (√1 - 2)/(√1) = (1 - 2)/1 = - 1 → the curve passes through the point A (1 ; - 1)

    The function y looks like (u/v), so the derivative looks like: [(u'.v) - (v'.u)]/v² → where:

    u = √x - 2 = x^(1/2) - 2 → u' = (1/2) * 1 * x^[(1/2) - 1] = (1/2) * x^(- 1/2)

    v = ³√x = x^(1/3) → v' = (1/3) * 1 * x^[(1/3) - 1] = (1/2) * x^(- 2/3)

    f'(x) = [(u'.v) - (v'.u)]/v²

    f'(x) = [{(1/2) * x^(- 1/2) * x^(1/3)} - {(1/2) * x^(- 2/3) * (√x - 2)}] / [x^(1/3)]²

    f'(x) = (1/2).[{x^(- 1/2) * x^(1/3)} - {x^(- 2/3) * (√x - 2)}] / x^(2/3)

    f'(x) = (1/2).[x^(- 1/6) - (√x - 2).x^(- 2/3)] / x^(2/3) ← this is the derivative

    …but the derivative is too the slope of the tangent line to the curve at x

    f'(x) = (1/2).[x^(- 1/6) - (√x - 2).x^(- 2/3)] / x^(2/3) → when: x = 1 (abscissa of A)

    f'(x) = (1/2).[1^(- 1/6) - (√1 - 2).1^(- 2/3)] / 1^(2/3)

    f'(x) = (1/2).[1 - (- 1).1] / 1

    f'(x) = (1/2).[2]

    f'(x) = 1 ← this is the slope of the tangent line to the curve at point A (1 ; - 1)

    The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

    The slope of the tangent line is (1), so the equation of this tangent line is: y = x + y₀

    y = x + y₀

    y₀ = y - x → you substitute x and y by the coordinates of the point A (1 ; - 1)

    y₀ = - 1 - 1

    y₀ = - 2

    The equation of the tangent line to the curve at A (1 ; - 1) is: y = x - 2

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  • Philip
    Lv 6
    3 weeks ago

    y = [x^(1/2) -2]/x^(1/3) = x^{1/2 - 1/3} - 2x^(-1/3) = x^(1/6) - 2x^(-1/3);

    (dydx) = (1/6)x^(-5/6) + (2/3)x^(-4/3) = (1/6)x^(-5/6)[1+4x^(-1/2)]. Now, at P(1,-1), y' =

    (1/6)5 = (5/6). Any line passing through P has slope = (5/6). We can adopt the slope,

    y-intercept form of equation of any line, L, say. L's equation is y = (5/6)x +b. Now.;

    since P is on L, we have  -1 =(5/6)1 +b. Then b =- 1 5/6 =(-11/6) & y = (5/6)x -(11/6),

    ie., 6y = 5x - 11, ie., 5x -6y -11 = 0.

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  • 1 month ago

    Use the Quotient Rule

    dy/dx = (vdu - udv) / v^2

    Also remember the 'square root' symbol can be written as a 'power of '1/2''

    Hence dy/dx = {[3x^(1/2)][(1/2)x^(-1/2)] - [x^(1/2) - 2 )((1/2)3x^(-1/2)]} / (9x )

    dy/dx = {[(3/2) ] - [3/2 - 3x^(-1/2)]} / 9x

    dy/dx = { 3/x^(1/2)] / 9x

    dy/dx = 1/ 3x^(3/2)

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  • 1 month ago

    You don't need the quotient rule here.  That's

    y = [x^(1/2) - 2] / x^(1/3) = x^(1/2 - 1/3)  - 2x^(-1/3)

    y = x^(1/6) - 2x^(-1/3)

    Now its two simple power-rule derivatives:

    y' = (1/6) x^(-5/6) - 2(-1/3) x ^(-4/3)

    y' = [x^(-5/6) + 4 x^(-4/3)] / 6

    You can simplify that a bit more by multiplying top and bottom by x^(4/3):

    y' = [x^(1/2) + 4] / [6 x^(4/3)]

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  • 1 month ago

    y = (√x -2) / (3√x)

    quotient rule 

    (f/g)' = (f'g–fg')/g²

    f = (√x -2)

    f' = 1/(2√x)

    g = (3√x)

    g' = 3/(2√x)

    g² = 9/x

    y' = ((3√x)/(2√x) – 3(√x -2)/(2√x) ) / 9/x

    y' = (3x/9•2)( (1) – (√x -2)/(√x) )

    y' = (x/6)( 1 – (√x)/(√x) + (2)/(√x) )

    y' = (x/6)( (2)/(√x) )

    y' = (x/3√x )

    y' = √x/3

    but check the math

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