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Math Help - Derivatives HELP!!!

  1. #1
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    Derivatives HELP!!!

    I have a few derivative problems that need computation, they are as follows...

    1. Find second derivative: y=(x^2-5)*e^x

    Find Derivatives:

    1. f(x)=\frac{3-x^4}{x^2+5}

    2. f(x)=(14-3x^2)^5

    3. f(x)=\sqrt{x^2-3}

    4. g(x)=2Ln(x^5)

    5. h(x)=5x^2*LnX
    ------------------------------------------------------
    My answers:
    1. Find second derivative:
    y1=2x*e^x+(x^2-5)*e^x
    y2=e^x+2*e^x

    Find Derivatives:
    1. f1(x)=\frac{(-4x^3)(x^2+5)-(3-x^4)(2x)}{(x^2+5)^2}
    = f1(x)=\frac{(-4x^3)-(3-x^4)(2x)}{(x^2+5)}


    2. f1(x)= -6x*5u^4
    = f1(x)=-6x*5(14-3x^2)^4
    = f1(x)=-6x*(70-15x^2)^4
    .........IDK Let u=14-3x^2 Let y=u^5

    3. confused

    4. g1(x)=
    Let u=x^5 Let y=2LnU
    ....???


    5. h1(x)=10x*Lnx+5x^2*\frac{1}{x}
    = h1(x)=10x*Lnx+5x
    Last edited by MathKidd; March 26th 2010 at 09:55 PM.
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  2. #2
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    What have you tried to do so far?
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  3. #3
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    Tried studying the rules...
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  4. #4
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    Let us, the forum, see your attempts of solving for the derivatives so we can see where you may have erred.
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    Let us, the forum, see your attempts of solving for the derivatives so we can where you may have erred.
    If you're not sure where to start....there are basically three rules.
    If it is a product - use the product rule.
    If it is a quotient - use the quotient rule.
    if it is a composite function - use the chain rule.
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  6. #6
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    "if it is a composite function - use the chain rule."

    how to tell if its a composite function and what exactly is the chain rule??
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    Quote Originally Posted by MathKidd View Post
    "if it is a composite function - use the chain rule."

    how to tell if its a composite function and what exactly is the chain rule??
    A composite function has an "inner function" usually in brackets and then an outer function. Eg your question 2 and 3.

    Chain Rule:
    If y is a function of u and u is a function of x then dy/dx = dy/du * du/dx.

    In Q2: Let u = 14 - 3x^2 (inner function)
    and then y = u^5 (outer function)

    Then work out du/dx (easy) and dy/du (easy again)
    and multiply them together.
    Then replace u with 14 - 3x^2 in your final step.

    Try it for Q3...hint: Let u = x^2 -3
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  8. #8
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    Quote Originally Posted by MathKidd View Post
    I have a few derivative problems that need computation, they are as follows...

    1. Find second derivative: y=(x^2-5)*e^x

    Find Derivatives:

    1. f(x)=\frac{3-x^4}{x^2+5}

    2. f(x)=(14-3x^2)^5

    3. f(x)=\sqrt{x^2-3}

    4. g(x)=2Ln(x^5)

    5. h(x)=5x^2*LnX
    ------------------------------------------------------
    My answers:
    1. Find second derivative:
    y1=2x*e^x
    y2=1*e^x

    Find Derivatives:
    1. f1(x)=\frac{(-4x^3)(x^2+5)-(3-x^4)(2x)}{(x^2+5)^2}

    2. ????

    3. ????

    4. g1(x)=2Ln*x^5+2Ln*5x^4 ???!!!! (i dont know the derivative of 2Ln)

    5. h1(x)=10x*Lnx+5x^2*\frac{1}{x}
    Your solution to your first question is incorrect. You need to remember that "The derivative of a product is NOT the product of the derivatives"...that's where you went wrong...use the PRODUCT RULE.

    Q1 is correct but you could tidy up the top with a bit of algebra.
    Q2&3 see my post on Chain Rule
    Q4 Also use the Chain Rule here: let u = x^5
    Q5 WELL DONE!! Just multiply out your last term to get 5x.
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  9. #9
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    Still a tad bit confused with the chain rule....are my other problems cleaner?? and is my solution to the first problem is correct now?
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  10. #10
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    Quote Originally Posted by MathKidd View Post
    I have a few derivative problems that need computation, they are as follows...

    1. Find second derivative: y=(x^2-5)*e^x

    Find Derivatives:

    1. f(x)=\frac{3-x^4}{x^2+5}

    2. f(x)=(14-3x^2)^5

    3. f(x)=\sqrt{x^2-3}

    4. g(x)=2Ln(x^5)

    5. h(x)=5x^2*LnX
    ------------------------------------------------------
    My answers:
    1. Find second derivative:
    y1=2x*e^x+(x^2-5)*e^x
    y2=e^x+2*e^x

    Find Derivatives:
    1. f1(x)=\frac{(-4x^3)(x^2+5)-(3-x^4)(2x)}{(x^2+5)^2}
    = f1(x)=\frac{(-4x^3)-(3-x^4)(2x)}{(x^2+5)}


    2. f1(x)= -6x*5u^4
    = f1(x)=-6x*5(14-3x^2)^4
    = f1(x)=-6x*(70-15x^2)^4
    .........IDK Let u=14-3x^2 Let y=u^5

    3. confused

    4. g1(x)=
    Let u=x^5 Let y=2LnU
    ....???


    5. h1(x)=10x*Lnx+5x^2*\frac{1}{x}
    = h1(x)=10x*Lnx+5x
    You can check your solutions here: Wolfram|Alpha
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  11. #11
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    Help! ! !
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  12. #12
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    Quote Originally Posted by MathKidd View Post
    Help! ! !
    Product Rule:

    \frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

    This is used when you have two "things" with x multiplied together. Eg: x^2e^x

    \frac{d}{dx} x^2e^x = (x^2)'e^x + x^2(e^x)' = 2xe^x + x^2e^x

    ==================================


    Quotient Rule:

    \frac{d}{dx} \left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

    An extension of the product rule, this is used when you have a division of x terms.

    For example \frac{sin(x)}{x}

    \frac{d}{dx} \left(\frac{sin(x)}{x}\right) = \frac{xcos(x)-1sin(x)}{x^2}


    =======================================

    Product Rule:

    \frac{d}{dx}\,f[g(x)] = f'[g(x)]g'(x)

    Although it looks bad it isn't as bad on paper. It's used when you have a function of x inside another function.

    eg: \tan(x^2)

    \frac{d}{dx} \tan(x^2) = \frac{d}{dx}\tan(x^2) \cdot \frac{d}{dx}(x^2) = 2x\sec^2(x^2)
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  13. #13
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    I know those rules ALREADY!, I need help finding the 2nd derivative of solution one, and answering questions 2, 3, and 4.....in regards to the chain rule! I have know idea how to solve them....................
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  14. #14
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    Well I can help you with 3, what helps me is to put the square root as (1/2) so it would be (x^2-3)^(1/2)
    so then use the chain rule
    (1/2)*((x^2-3)^(-1/2))*(2x)
    hope this helps
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  15. #15
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    Quote Originally Posted by MathKidd View Post
    I have a few derivative problems that need computation, they are as follows...

    1. Find second derivative: y=(x^2-5)*e^x

    Find Derivatives:

    1. f(x)=\frac{3-x^4}{x^2+5}

    2. f(x)=(14-3x^2)^5

    3. f(x)=\sqrt{x^2-3}

    4. g(x)=2Ln(x^5)

    5. h(x)=5x^2*LnX
    ------------------------------------------------------
    My answers:
    1. Find second derivative:
    y1=2x*e^x+(x^2-5)*e^x
    y2=e^x+2*e^x

    Find Derivatives:
    1. f1(x)=\frac{(-4x^3)(x^2+5)-(3-x^4)(2x)}{(x^2+5)^2}
    = f1(x)=\frac{(-4x^3)-(3-x^4)(2x)}{(x^2+5)}


    2. f1(x)= -6x*5u^4
    = f1(x)=-6x*5(14-3x^2)^4
    = f1(x)=-6x*(70-15x^2)^4
    .........IDK Let u=14-3x^2 Let y=u^5

    3. confused

    4. g1(x)=
    Let u=x^5 Let y=2LnU
    ....???


    5. h1(x)=10x*Lnx+5x^2*\frac{1}{x}
    = h1(x)=10x*Lnx+5x
    In First question, your first derivative is correct.
    Now you need to use the product rule on each term to find the second derivative. So find the derivative of 2xe^x first (using product rule) and then find the derivative of (x^2-5)e^x (using the product rule again). Then add them together and tidy up.

    Q1 You differentiated correctly (using the quotient rule) but then stuffed up the tidying up. You can't cancel (x^2-5) because it is not a factor of the top. The best you can do is expand the brackets (on the top) and collect like terms. Leave the bottom as is.

    Q2 Your first 2 lines are correct - so you have applied the Chain Rule correctly. Again you've stuffed up in the tidying up. (You know what - Calculus isn't your real problem - algebra is). You can take the 5 into the bracket because the bracket is raised to the power 4 and the 5 isn't. You should just get -30x(14-3x^2)^4.

    Q3 Write it as (x^2-3)^0.5 (because square root is the same as power of 1/2) and then do it the same way as Q2 (Chain rule).

    Q4 You have set up question 4 correctly. Now find du/dx (ie differentiate x^5 with respect to x) and find dy/du (ie differentiate 2 ln u with respect to u). Then muliply them together (that's the chain rule). You know the derivative of ln u because you have differentiated a ln function in Q5.

    Stick with it...you're nearly there!!
    Last edited by Debsta; March 27th 2010 at 02:42 PM.
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