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Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Find the derivative of the function using the definition of derivative. f(x) = 2x^4?

8 Answers

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  • 2 months ago
    Favorite Answer

    f(x + h) = 2(x + h)^4 = 2(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4)

                 = 2x^4 + 8x^3h + 12x^2h^2 + 8xh^3 + 2h^4

    f(x + h) - f(x) = f(x + h) = 2x^4 = 8x^3h + 12x^2h^2 + 8xh^3 + 2h^4

    [f(x + h) - f(x)] / h = 8x^3 + 12x^2h + 8xh^2 + 2h^3

    f'(x) = lim h-->0 [f(x + h) - f(x)] / h = 8x^3

    All the other terms have a factor of h and vanish as h-->0.

  • 2 months ago

    the answer is 8x^3 because

    2x^4

    bring 4 to the front so:

    4(2)x^

    is 8x^ now:

    now the exponent you subtract one from all exponents:

    4-1=3 so

    8x^3

  • 2 months ago

    f(x) = 2x^4 

    f'(x) = 4(2)x^(4-1) 

    f'(x) = 8x^3 

  • 2 months ago

    The definition of a derivative is:

    . . . . . f(x + h) - f(x)

    lim . . -----------------

    h->0 . . . . . h

    You have:

    f(x) = 2x^4

    You can also figure out that:

    f(x + h) = 2(x + h)^4

    The hard part is to now expand that binomial raised to the 4th power. I simply used the expansion for (a + b)^n = C(n,0) a^n b^0 + C(n,1) a^(n-1) b^1 + ... + C(n,n) a^0 b^n.

    Or you can expand it all out by distributing. 

    f(x + h) = 2(x + h)(x + h)(x + h)(x + h)

    f(x + h) = 2(x² + 2hx + h²)(x² + 2hx + h²)

    f(x + h) = 2[x²(x² + 2hx + h²) + 2hx(x² + 2hx + h²) + h²(x² + 2hx + h²)]

    etc.

    However you do it, you'll get to:

    f(x + h) = 2[x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4]

    f(x + h) = 2x^4 + 8x^3h + 12x^2h^2 + 8xh^3 + 2h^4

    When you subtract f(x), the first term disappears:

    f(x + h) - f(x) = 8x^3h + 12x^2h^2 + 8xh^3 + 2h^4

    Next divide it by h:

    [f(x + h) - f(x)]/h = 8x^3 + 12x^2h + 8xh^2 + 2h^3

    When you take the limit as h approaches 0, the last 3 terms disappear.

    Answer:

    8x^3

  • 2 months ago

    f '(x)=

    limit {2[(x+h)^4-x^4]/h}, where h is a small increment in x.

    h->0

    =

    limit {2[(x+h)^2+x^2][(x+h)^2-x^2]/h}

    h->0

    =

    limit {2[(x+h)^2+x^2][2hx+h^2]/h}

    h->0

    =

    2[x^2+x^2][2x]

    =

    8x3

  • 2 months ago

    f(x) = 2 x^4

    d/dx(2 x^4) = 8 x^3

  • ?
    Lv 7
    2 months ago

    Recall that the definition is [f(x + h) - f(x)] / h  

    as h approaches zero; it's a limit.

    So, what if f(x) ?

    What is f(x+h).

    Subtracting these & simplifying gives your numerator. 

    Divide it all now by h & simplify.

    What do you get?

    Then evaluate this as h gets very small.

    That's your answer! Do do these steps!

    Don't forget to vote me best answer for being the first to correctly walk you through without spoiling the answer. That way it gives you a chance to work at it and to get good at it!

  • ?
    Lv 7
    2 months ago

    f(x) = 2 x^4

    dy/dx  =  2 * 4  x^(4-1)  =  8 x^3 ............ Answer

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