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Math Help - Differentiation

  1. #1
    DRJ
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    Differentiation

    I am currently studying calculus on an open learning basis and was wondering if someone could take a look at these questions and answers I have posted and feed back on my method of working out etc. I am pretty sure these answers are correct but would appreciate any comments.

    Question 1:

    Obtain \frac{dy}{dx} for the following expression.

    y=(5x + 4)^3 so 3(5x + 4)^2 This is the 1st part.

    I then differentiate the inner part of the function so:-

    (5x + 4) differentiates to 5. This is the the 2nd part.

    I then multiply the 1st by the 2nd to get 3(5x + 4)^2\times 5

    so \frac{dy}{dx}=15(5x + 4)^2


    Question 2:

    Obtain \frac{dy}{dx} for the following expression.

    y=(3 - 2x)^5 so 5(3 - 2x)^4 This is the 1st part.

    I then differentiate the inner part of the function so:-

    (3 - 2x) differentiates to -2. This is the the 2nd part.

    I then multiply the 1st by the 2nd to get 5(3 - 2x)^4\times -2

    so \frac{dy}{dx}=-10(3 - 2x)^4


    Question 3:

    Obtain \frac{dy}{dx} for the following expression.

    y=\sqrt{(5 - 0.6x)} which is  (5 - 0.6x)^\frac{1}{2}

    I then differentiate this to get \frac{1}{2}(5 - 0.6x)^{-\frac{1}{2}} This is the 1st part:

    (5 - 0.6x) differentiates to -0.6. This is the the 2nd part.

    I then multiply the 1st by the 2nd to get \frac{1}{2}(5 - 0.6x)^{-\frac{1}{2}}\times -0.6

    so \frac{dy}{dx}=-0.3(5 - 0.6x)^{-\frac{1}{2}}


    Question 4:

    Obtain \frac{dy}{dx} for the following expression.

    y=(2 + 3x)^-0.6 so -0.6(2 + 3x)^{-1.6} This is the 1st part.

    I then differentiate the inner part of the function so:-

    (2 + 3x) differentiates to 3. This is the the 2nd part.

    I then multiply the 1st by the 2nd to get -0.6(2 + 3x)^{-1.6}\times 3

    so \frac{dy}{dx}=-1.8(2 + 3x)^{-1.6}

    As I said at the beginning I am pretty sure I am 100% correct and I agree these are relatively simple questions from a calculus point of view but being new to this subject I was just wondering if someone could make comment please.

    Thanks

    Derek.
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  2. #2
    Lord of certain Rings
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    I think they are all right

    And there is nothing much to comment either... Why not try harder ones??
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  3. #3
    DRJ
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    Obtain the derivative of

    I came across another area which I am trying to get to grips with and was wondering if someone could help

    I need to obtain the derivative of f(t) = ln(5 - \frac{2}{3}t)

    I think I procede as follows.

    \frac{dy}{dt} = \frac{1}{5 - \frac{2}{3}t}\times -\frac{2}{3}

    Finishing up with f'(t) = \frac{-\frac{2}{3}}{5 - \frac{2}{3}t}

    or \frac{-0.6}{5 - 0.6 t}

    Another one is Determine f'(y)if f(y) = exp(3 - \frac{1}{4}y)

    When the question reads "exp" do I assumeit is refering to the exponential function e

    I have hunted high and low for similar equations to this to give me some kind of example to work too but come up with more or less nothing so from what I have found I have tried with this. I am probably wrong. Could someone make comment please??

    Thanks

    Derek
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by DRJ View Post
    Another one is Determine f'(y)if f(y) = exp(3 - \frac{1}{4}y)

    When the question reads "exp" do I assumeit is refering to the exponential function e

    I have hunted high and low for similar equations to this to give me some kind of example to work too but come up with more or less nothing so from what I have found I have tried with this. I am probably wrong. Could someone make comment please??

    Thanks

    Derek
    If f(x) = e^{ax+b} then f'(x) = ae^{ax+b}
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DRJ View Post
    I came across another area which I am trying to get to grips with and was wondering if someone could help

    I need to obtain the derivative of f(t) = ln(5 - \frac{2}{3}t)

    I think I procede as follows.

    \frac{dy}{dt} = \frac{1}{5 - \frac{2}{3}t}\times -\frac{2}{3}

    Finishing up with f'(t) = \frac{-\frac{2}{3}}{5 - \frac{2}{3}t}

    or \frac{-0.6}{5 - 0.6 t}
    Yes, though if this is a Math class I'd recommend leaving this as
    f^{\prime}(t) = -\frac{2}{15 - 2t}
    rather than the decimal form.

    Quote Originally Posted by DRJ View Post
    When the question reads "exp" do I assumeit is refering to the exponential function e
    Yes, the "exp" and "e" exponential are the same function.

    -Dan
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  6. #6
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    Hello, DRJ!

    Find the derivative of: f(t) \:= \:\ln\left(5 - \frac{2}{3}t\right)

    I think I proceed as follows: . \frac{dy}{dt} \:= \:\frac{1}{5 - \frac{2}{3}t} \times \left(-\frac{2}{3}\right)

    Finishing up with: . f'(t) \:= \:\frac{-\frac{2}{3}}{5 - \frac{2}{3}t} . . . . Yes!

    . . or: . \frac{-0.6}{5 - 0.6 t} . . . . no
    Don't use rounded-off decimals . . . 0.6 is not equal to \frac{2}{3}

    We have: . f'(x) \:=\:\frac{-\frac{2}{3}}{5 - \frac{2}{3}t}

    Multiply top and bottom by 3: . \frac{3 \times\left(-\frac{2}{3}\right)}{3 \times\left(5 - \frac{2}{3}t\right)} \;=\;\frac{-2}{15-2t}



    Determine f'(y)\text{ if }f(y) \:= \:\text{exp}\left(3 - \frac{1}{4}y\right)

    When the question reads "exp",
    do I assume it is refering to the exponential function e ? . Yes!
    . . . . . . . . . . . . . . the original function
    . . . . . . . . . . . . . . . . . . . \downarrow
    Given: . y \:=\:e^u, then: . y' \:=\:e^u\cdot u'
    . - . - . - . . . . . . . . . . . . . . . \uparrow
    . . . . . . . . . . . . . . .
    derivative of the exponent

    We have: . f(y) \:=\;e^{(3-\frac{1}{4}y)}

    Then: . f'(y) \;=\;e^{(3-\frac{1}{4}y)}\!\cdot\!\left(\text{-}\frac{1}{4}\right) \;=\;-\frac{1}{4}e^{(3-\frac{1}{4}y)}

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  7. #7
    DRJ
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    Differentiation and the exponential function

    Hello again

    Just a quick query regarding the e function.

    In my example f(y) = exp(3 - \frac{1}{4}y) I think I may have mis read the question.

    Is this actually asking me to find f(y) of exp^{(3-\frac{1}{4}y)} or as you stated e^{(3-\frac{1}{4}y)} as opposed to f(y) = exp(3 - \frac{1}{4}y) or are they both the same???

    Thanks and happy new year etc
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DRJ View Post
    Hello again

    Just a quick query regarding the e function.

    In my example f(y) = exp(3 - \frac{1}{4}y) I think I may have mis read the question.

    Is this actually asking me to find f(y) of exp^{(3-\frac{1}{4}y)} or as you stated e^{(3-\frac{1}{4}y)} as opposed to f(y) = exp(3 - \frac{1}{4}y) or are they both the same???

    Thanks and happy new year etc
    There is no such thing as exp^x. exp(x) is the operation e^x, so there is no room for confusion.

    -Dan
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