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Thread: Differentiation problems.

  1. #1
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    Differentiation problems.

    Differentiate the following:
    1. $\displaystyle f(x) =\sqrt{x}(2x+3)^{2} $
    2. $\displaystyle f(x) = \frac{sinx}{x^3-2x}$

    My attempt for 1:
    First I rewrite to get rid of the radical:
    $\displaystyle f(x) = x^{\frac{1}{2}}(2x+3)^{2}$
    Then I start differentiating using product rule:
    $\displaystyle f'(x) = x^{\frac{1}{2}}'(2x+3)^{2} +x^{\frac{1}{2}}(2x+3)^{2}' $
    Then chain rule:
    $\displaystyle f'(x) = (\frac{1}{2})(x^{\frac{1}{2}-1)}(1)(2x+3)^{2} +(x^{\frac{1}{2}})(2)(2x+3)^{(2-1)}(2)$
    $\displaystyle f'(x) = (\frac{1}{2})(x^{\frac{-1}{2})}(1)(2x+3)^{2}+(x^{\frac{1}{2}})(4)(2x+3) $
    After simplifying:
    $\displaystyle f'(x) = \frac{(2x+3)^{2}}{\sqrt{2}\sqrt{x}} + \frac{2x+3}{2\sqrt{x}}$

    My attempt for number 2:
    First I rewrite the equation to get rid of the fraction:
    $\displaystyle f(x) = (sinx)(x^3-2x)^{-1)$
    Then I begin to differentiate using product rule:
    $\displaystyle f'(x) = (sinx)'(x^3-2)^{-1} +(sinx)(x^3-2x)^{-1}'$
    Chain rule:
    $\displaystyle (cosx)(x^3-2x)^{-1} + (sinx)(-1)(x^3-2x)^{-1-1}(3x^2-2)$
    After simplifying:
    $\displaystyle f'(x) = \frac{cosx}{(x^3-2x)} + \frac{2(3x^2 sinx)}{(x^3-2x)^{2}}$

    What have I done wrong in my calculations? To check that I'm correct, I plug in random and find the slopes of the tangent to those points and check if the functions have been derived correctly through a graphing program. Unfortunately, I am way off. Thanks for any help in advance.
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  2. #2
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    Hello, Pupil!

    You made silly mistakes in the last step . . .


    Differentiate the following:

    $\displaystyle 1.\;f(x) \:=\:\sqrt{x}(2x+3)^{2} $

    $\displaystyle 2.\;f(x) \:=\:\dfrac{\sin x}{x^3-2x}$



    My attempt for 1:

    First I rewrite to get rid of the radical: .$\displaystyle f(x) \:=\: x^{\frac{1}{2}}(2x+3)^{2}$


    Then I start differentiating using product rule:
    $\displaystyle f'(x) \:=\: x^{\frac{1}{2}}'(2x+3)^2 +x^{\frac{1}{2}}(2x+3)^2' $


    Then chain rule:
    $\displaystyle f'(x) \:=\: (\frac{1}{2})(x^{-\frac{1}{2}})(2x+3)^2+(x^{\frac{1}{2}})(4)(2x+3) $

    After simplifying:

    $\displaystyle f'(x) = \dfrac{(2x+3)^{2}}{\sqrt{2}\sqrt{x}} + \dfrac{2x+3}{2\sqrt{x}}$
    . . . . . . . . . . . . . . . . . . .
    . . . . . . . just 2. . . . . . in numerator



    My attempt for number 2:

    First I rewrite the equation to get rid of the fraction:
    $\displaystyle f(x) = (\sin x)(x^3-2x)^{\text{-}1}$

    Then I begin to differentiate using product rule:
    $\displaystyle f'(x) = (\sin x)'(x^3-2)^{\text{-}1} +(\sin x)(x^3-2x)^{\text{-}1}'$

    Chain rule:
    $\displaystyle (\cos x)(x^3-2x)^{\text{-}1} + (\sin x)(-1)(x^3-2x)^{\text{-}2}(3x^2-2)$


    After simplifying:. . . . . . .?
    . . . . . . . . . . . . . . . . . . .
    $\displaystyle f'(x) \:=\: \dfrac{\cos x}{(x^3-2x)} + \dfrac{2(3x^2-2)\sin x}{(x^3-2x)^{2}}$
    . . . . . . . . . . . . . . . . .
    . . . . . . . . . . . . . . . minus

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  3. #3
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    Hi Soroban, thanks for the reply.

    Okay, so for 1:
    $\displaystyle f'(x) = \frac{(2x+3)^{2}}{2\sqrt{x}} + \sqrt{x}(4)(2x+3)$ is this correct?

    And for 2:
    $\displaystyle f'(x) = \frac{cosx}{(x^3-2x)} - \frac{(3x^2 - 2sinx)}{(x^3-2x)^{2}}$
    ?
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