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Math Help - Help to find solutions

  1. #1
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    Help to find solutions

    1. suppose that f(x) and g(x) are twise differentiable at x=a,
    show that
    (fg)"(a)= f(a) g"(a) + 2f'(a) g'(a) + f"(a) g(a)
    __________________________________________________ _____________

    2. Use the definition of derivative to show that for the function

    f(x)=3(x)^(1/3)

    (f of x is equal to 3 x to the power 1/3)

    we have f'(a)=a^(2/3), for every "a" belongs to "R" (Set of real numbers)

    [Use the identity (a^3)-(b^3)= (a-b)((a^2)+ab+(b^2))]
    __________________________________________________ _____________

    3. Differentiate the following and prove the necessary expressions as given

    a) If y=xtanx

    show that

    x sinx (dy/dx)=xtanx +y sinx
    _____
    b) y= (x+√ x-1 )ⁿ
    prove that

    (x-1) (dy/dx)=ny


    Plz help me to solve these.
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  2. #2
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    nobody
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  3. #3
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    Quote Originally Posted by chanaka89 View Post
    2. Use the definition of derivative to show that for the function

    f(x)=3(x)^(1/3)

    (f of x is equal to 3 x to the power 1/3)

    we have f'(a)=a^(2/3), for every "a" belongs to "R" (Set of real numbers)

    [Use the identity (a^3)-(b^3)= (a-b)((a^2)+ab+(b^2))]
    _________________________________________________
    f(x) = 3x^{\frac{1}{3}}.


    f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

     = \lim_{h \to 0}\frac{3(x + h)^{\frac{1}{3}} - 3x^{\frac{1}{3}}}{h}

     = \lim_{h \to 0}\frac{3[(x + h)^{\frac{1}{3}} - x^{\frac{1}{3}}]}{h}

     = \lim_{h \to 0}\frac{3\left[\frac{(x + h) - x}{(x + h)^{\frac{2}{3}} + + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}\right]}{h}

     = \lim_{h \to 0}\frac{\frac{3h}{(x + h)^{\frac{2}{3}} + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} +  x^{\frac{2}{3}}}}{h}

     = \lim_{h \to 0}\frac{3}{(x + h)^{\frac{2}{3}} + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} +   x^{\frac{2}{3}}}

     = \frac{3}{x^{\frac{2}{3}} + x^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}

     = \frac{3}{3x^{\frac{2}{3}}}

     = \frac{1}{x^{\frac{2}{3}}}

     = x^{-\frac{2}{3}}
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  4. #4
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    Smile

    Quote Originally Posted by Prove It View Post
    f(x) = 3x^{\frac{1}{3}}.


    f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

     = \lim_{h \to 0}\frac{3(x + h)^{\frac{1}{3}} - 3x^{\frac{1}{3}}}{h}

     = \lim_{h \to 0}\frac{3[(x + h)^{\frac{1}{3}} - x^{\frac{1}{3}}]}{h}

     = \lim_{h \to 0}\frac{3\left[\frac{(x + h) - x}{(x + h)^{\frac{2}{3}} + + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}\right]}{h}

     = \lim_{h \to 0}\frac{\frac{3h}{(x + h)^{\frac{2}{3}} + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} +  x^{\frac{2}{3}}}}{h}

     = \lim_{h \to 0}\frac{3}{(x + h)^{\frac{2}{3}} + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} +   x^{\frac{2}{3}}}

     = \frac{3}{x^{\frac{2}{3}} + x^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}

     = \frac{3}{3x^{\frac{2}{3}}}

     = \frac{1}{x^{\frac{2}{3}}}

     = x^{-\frac{2}{3}}
    Thanks
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  5. #5
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    Quote Originally Posted by chanaka89 View Post
    1. suppose that f(x) and g(x) are twise differentiable at x=a,
    show that
    (fg)"(a)= f(a) g"(a) + 2f'(a) g'(a) + f"(a) g(a)
    Use the product rule- twice.

    __________________________________________________ _____________

    2. Use the definition of derivative to show that for the function

    f(x)=3(x)^(1/3)

    (f of x is equal to 3 x to the power 1/3)

    we have f'(a)=a^(2/3), for every "a" belongs to "R" (Set of real numbers)

    [Use the identity (a^3)-(b^3)= (a-b)((a^2)+ab+(b^2))]
    __________________________________________________ _____________

    3. Differentiate the following and prove the necessary expressions as given

    a) If y=xtanx

    show that

    x sinx (dy/dx)=xtanx +y sinx
    Okay, what did you get for dy/dx?
    _____
    b) y= (x+√ x-1 )ⁿ
    prove that

    (x-1) (dy/dx)=ny
    Again, what did you get for dy/dx?

    Plz help me to solve these.
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