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Math Help - 2 Questions Derivative

  1. #1
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    2 Questions Derivative

    Two questions:

    #1:



    (fog)'(3)=?
    (gof)'(3)=?

    #2:

    Let F(x)= f(x^{6}) and G(x)=(f(x))^{6} . You also know that a^{5}=10, f(a)=2,f'(a)=12, f'(a^{6})=10.

    Find F'(a)=
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Two questions:

    #1:



    (fog)'(3)=?
    (gof)'(3)=?

    #2:

    Let F(x)= f(x^{6}) and G(x)=(f(x))^{6} . You also know that a^{5}=10, f(a)=2,f'(a)=12, f'(a^{6})=10.

    Find F'(a)=
    #1 I'd like to help, but I'm not sure what the notation (fog)'(3) means. Is the ' a prime, representing a derivative?

    #2 Find F'(a):
    Since F(x) = f(x^6), taking the derivative, we need to use chain rule:
    F'(x) = f'(x^6)*6x^5
    F'(a) = f'(a^6)*6a^5

    Now, we know f'(a^{6})=10, a^{5}=10, so we get
    F'(a) = {10}*6*{10} = 600
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  3. #3
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    Re:

    F composed of g like a composition...



    Also I have another question...


    If I wanted to differentiate

    -2y

    for implicit differentiation would the derivative be

    -2 or

    -2yy^prime




    Thanks so much
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    F composed of g like a composition...



    Also I have another question...


    If I wanted to differentiate

    -2y

    for implicit differentiation would the derivative be

    -2 or

    -2yy^prime




    Thanks so much
    -2 y'

    that is if you're differentiating with respect to a variable other than y
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by qbkr21 View Post
    F composed of g like a composition...



    Also I have another question...


    If I wanted to differentiate

    -2y

    for implicit differentiation would the derivative be

    -2 or

    -2yy^prime




    Thanks so much
    d/dx(-2y) = -2*dy/dx = -2*y' <-- the ' is "prime"
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    Re:

    Thanks!
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  7. #7
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    Re:

    I am in Calc I. Typically is there just 1 y' for ever derivation. For instance I mean:

    y^3 so 3y^2y'

    another

    2y^2 so 4yy'

    again

    -2y so -2y'

    is this correct... if so I am set!!!
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  8. #8
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Two questions:

    #1:



    (fog)'(3)=?
    (gof)'(3)=?

    Find F'(a)=
    (fog)(3) = f(g(3))
    First we need the value of g at x=3: g(3) = 4. To get this, just look on the x-axis where x=3 and move up until you hit the graph of g, the y-value at that point is g(3)=4.
    Second, we need f(g(3)) = f(4) = 2. Use the same method for x=4, going up to the graph of f. The y-value at that point is f(4)=2.

    Now, do the same for (gof)(3) = g(f(3)).
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    I am in Calc I. Typically is there just 1 y' for ever derivation. For instance I mean:

    y^3 so 3y^2y'

    another

    2y^2 so 4yy'

    again

    -2y so -2y'

    is this correct... if so I am set!!!
    correct
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    (fog)(3) = f(g(3))
    First we need the value of g at x=3: g(3) = 4. To get this, just look on the x-axis where x=3 and move up until you hit the graph of g, the y-value at that point is g(3)=4.
    Second, we need f(g(3)) = f(4) = 2. Use the same method for x=4, going up to the graph of f. The y-value at that point is f(4)=2.

    Now, do the same for (gof)(3) = g(f(3)).
    so we're forgetting about the primes? i hope we can do that
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  11. #11
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    Re:

    Thanks you guys. YOU ARE THE BEST!!!
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  12. #12
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    Re:

    Still confused a bit

    (f composed of g)' is not equal to 2 according to the computer...
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  13. #13
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by qbkr21 View Post
    I am in Calc I. Typically is there just 1 y' for ever derivation. For instance I mean:

    y^3 so 3y^2y'

    another

    2y^2 so 4yy'

    again

    -2y so -2y'

    is this correct... if so I am set!!!
    The technical rule is that any time you 'take the derivative of y' you get a y'. For example, let's say I'm taking the derivative of 2xy implicitly. This is product rule, so: D(2xy) = 2x*y' + 2y. When I took the derivative of y, I got y', when I took the derivative of x, 1.

    Besides the technical rule, what you have done is correct.
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  14. #14
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jhevon View Post
    so we're forgetting about the primes? i hope we can do that
    That's why I asked about the ' . I'm not sure what to do with the prime there.
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  15. #15
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    Re:

    Does anyone know how to solve question #1 with the graph? After studying the picture for quite some time I have come to the conclusion that the lines on the graph are f(x) and g(x), they are not f'(x) and g'(x). Any assistance would wonderful. This is my last problem for Webwork and I am clueless as to what I should do.

    Thanks!
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