Derivatives in calculus?

hi, i'm 14 and i'm teaching myself how to do derivatives and i've been pretty successful so far e.g

f(x)=5x^5+6x^4+3x^2+3

f '(x)=25x^4+24x^3+3x

i'm pretty sure i got that right anyway is their more to derivatives than stuff like that

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  • 1 decade ago
    Best Answer

    yes. Those are for simple functions involving cx^n

    There are derivative rules like:

    1) function of a function/chain/composite rule

    If y = f(x)^n , then dy/dx = n f(x)^(n-1) . f '(x) or dy/dx = du/dx . dy/du

    for instance differentiate (x^2 + 3)^3

    dy/dx = 3(x+3)^2 . 2x

    = 6x(x+3)^2

    really chain rule derivative is the product of the derivative of the function inside the brackets and the derivative of the whole function.

    2)Product rule

    If y = uv, then y' = uv' + vu'

    example: y = 2x(x^3 - 5x)

    Let u = 2x and v = (x^3 - 5x)

    Then uv' = 2x(3x^2 - 5) and vu' = (x^3 - 5x)(2)

    y' = (6x^3 - 10x) + (2x^3 - 10x)

    = 8x^3 - 20x

    3) Quotient rule:

    If y = u/v, then dy/dx = (vu' - uv') / v^2

    there are also physical applications of calculus, tangents and normals, rates of change, points of inflection, primitive functions, second derivatives, other derivatives for trigonometry, eg. y = sinx, then y' = cos x, logarithms and natural logs, but you won't need to know these as they are HSC topics. I know you will have fun. good on you. You will look forward to integration too, when you get there...

  • 1 decade ago

    You were close the last variable in f '(x) should be 6x but that was probably just a typo, thats pretty impressive if you really are 14 cuz i def wasnt doin derivatives then haha.

    Yea man there's loads more to do with derivatives, those are just the very basic derivatives you learn the idea with, then you go in to using sin, cos, tan, etc. all of those. You do alot more complex derivatives and you learn different rules that go along with it, two of the popular ones would be the product and quotient rule. You do them backwards those are called antiderivatives or intervals(those are in calc 2), you do multivariable deferintiation which is pretty much doing multiple derivates or intervals at once and all that it gets pretty difficult, that stuff is in calc 3, and that's as far as I've gotten so idk whats after that. But after you get used to it, it's actually pretty cool. You can use them to find acceleration, velocity, and speed or position, which is helpful in physics too.

  • Rachel
    Lv 4
    1 decade ago

    yes, unfortunately.

    But you've got finding the derivative of polynomials down. That's great that you're teaching yourself.

    If you look online, there are some derivative lists that it is helpful to have memorized. If you look enough you can find proofs (showing the steps) of those derivatives which is helpful also.

    There is also higher order derivatives (like second, third, fourth derivatives), there are derivative rules for products and quotients, logarithmic/exponential derivatives, implicit differentiation.

    There's also applications of derivatives (how they are used in the real world)

    So there is a lot more, but really, congrats on tackling this on your own. Shows initiative, powerful work ethic, something most people your age seriously lack.

  • 1 decade ago

    There are some fundamental rules that make derivatives more useful, such as:

    if f(x) = a(x) * b(x)

    then

    f '(x) = a'(x)*b(x) + a(x)* b'(x)

    There are also specific derivatives that have broad applicability, but I'm not sure if you would have encountered the natural number e yet. e is a fundamental constant of nature and its value is roughly 2.71828. It is very important in complex number analysis, which has broad application in technology. One interesting property is that the derivative of "e to the "x" power", sometimes written as exp(x), is itself!

    In other words, if f(x) = exp(x), then f '(x) = exp(x) = f(x).

    You can read more about this "natural number" at the link below.

    I hope you continue to take initiative in educating yourself. It will pay off big time for the rest of your life!

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  • Anonymous
    1 decade ago

    yeah, calculus is all about derivatives. and derivatives get alot more complicated than that. wat u are learning about is the power rule. the power rule is pretty simple actually. if u have a function f(x) = cx^n, the derivative f '(x) = cnx^(n - 1). later, u will have to kno the product rule, the quotient rule, the chain rule, derivatives of log functions, derivatives of trig functions, derivatives of exponential functions, and most complicated of all is implicit differentiation. and if u havent learned limits yet, i would do so. it is important to understand the concept of a limit and how it applies to derivatives. if u dont kno limits, then youre not gonna get alot out of calculus. so yeah, u only learned a very little part of calculus. it gets MUCH harder than that, trust me. and btw, the last term of the derivative u calculated is supposed to be 6x, not 3x.

  • 1 decade ago

    To determine the derivative of a polynomial, multiply the exponent of each term by its coefficient, decrease the exponent by 1 and disregard any constants.

    In your example f '(x), the last term should be 6x.

    Then there are derivatives of sums, f(x) + g(x), differences, f(x) - g(x), products, f(x)g(x), quotients, f(x) / g(x), exponents, f(x)^n, and trig functions.

    If you are teaching yourself derivatives and understanding it, you have my admiration.

  • 3 years ago

    it ought to help to do a u substitution. y = (a million+a million/x)^3 permit u = a million + a million/x Then y' = f'(u)*u' y = u^3 y' = 3*u^2 * u' y' = 3* (a million + a million/x)^2 * (-a million/x^2) For y'' you will additionally want the product rule. y'' = 3*(a million+a million/x)^2 * (2/x^3) + (-a million/x^2)*by-product ( 3*(a million+a million/x)^2) utilising u substitution returned: -------------A------- y'' = [6*(a million+a million/x)^2]/x^3 -a million/x^2 * 6u*u' = A -a million/x^2* 6*(a million+a million/x)*-a million/x^2 = A + [6*(a million+a million/x)] / x^4

  • Anonymous
    1 decade ago

    5x^5->25x^4

    6x^4->24x^3

    3x^2->6x

  • 1 decade ago

    the lat term is 6x, not 3x

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