First Second
How do you find the first, second derivative?
How do you find the first, second derivative?
3x^(2/3)-x^2
Also critical numbers, asymptotes, and x and y intercepts
You can work derivatives out from first principles, ie. using the definition of what a derivative is; the rate of change of a function (basicly). But thats not really fun unless you are trying to proove something. What you will need to learn are the basic derivative ‘relationships’.
I’m just calling them relationships, they are derivatives that occur commonly, so, instead of using first principles (sometimes very hard) we just use the relationship someone else found ages ago between a function of a certain form f(x) and its derivative d/dx [f(x)]. I googled for a list of the basic ones, see my source.
Apply these to your function, using the same order as you would any algebra question (look at the brackets, then the multiply/divides, then the add/subtracts).
3x^(2/3) looks like its in the same form as the 3rd relationship under “Powers and polynomials” in the wiki link, so:
d/dx [ 3x^(2/3) ] = (2/3) * 3x^(2/3 – 1)
d/dx [ 3x^(2/3) ] = 2x^(-1/3) when we simplify
The second part is much easier:
d/dx [x^2] = 2x
and combining to find the total first derivative:
d/dx [ 3x^(2/3) - x^2 ] = 2x^(-1/3) – 2x
In case you are wondering:
d/dx [ f(x) ] ==> “derivative of the function f(x) with respect to x”
Now, the second derivative…. you just repeat, using the relationships on your first derivative:
d^2/dx^2 [ f(x) ] == d/dx [ d/dx {f(x)} ]
ie. The second derivitive of a function, is the derivative of the derivative.
so the second derivative is:
d/dx [ 2x^(-1/3) - 2x ] = [ (-1/3) * 2x^(-1/3 - 1) ] – 2
d/dx [ 2x^(-1/3) - 2x ] = (-2/3)x^(-4/3) – 2
Critical numbers: The value of x such that d/dx [ f(x) ] = 0. Ie. the x value for maximums, minimums, and those other funny s-shape turns (i forgot the name). For this, just set the first derivative to equal zero and solve:
0 = 2x^(-1/3) – 2x
2x = 2x^(-1/3)
x = x^(-1/3) … well x can only be 1, or -1.
Remember, the derivative is on the same axis! it is just the rate of change, these x-values are the values at which there is no change in the original function.
Asymptotes: Those random bits where the function just suddenly goes to infinite. You COULD find it by taking the limit of f(x) as x approaches infinite, but to be honest I just grabbed my graphics calculator and plotted it. It’s at 0… but read wiki for more info about working them out.
x-intercept: just take f(x) = 0 and solve it
y-intercept: just take f(0) and solve
… You asked a really basic calculus question, you’ll find it much easier to ask a teacher / tutor, then deciphering wordy yahoo! answers =P good luck… Buy a graphics calculator, haha, it helps check the sanity of your answers at least.
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