Understanding Derivative: Solving for f'(x) in Different Functions

[Paquita888]

Hi all

I am starting to learn Derivative,, does anybody wants to help me to show me

why does

f(x) = (x^2/ 3) - (3/x^2)
f'(x) = 2/3 x + 6/x^3 ?

and

f(x) = -3(2x^2 - 5x + 1)
f'(x) = 12x +15

I am really appreciate if anybody can show me how to get into the answer :) thanksss a lot!

 

[Fragment]

You will have to show work before anybody starts contributing a solution. Try using the limit definition of a derivative for simple cases,

f'(a)=\lim_{h\to 0}{f(a+h)-f(a)\over h}

Otherwise, you will need to know about the power rule.

 

[Inferior89]

f(x) = -3(2x^2 - 5x + 1)
f'(x) = 12x +15

This is wrong.

f'(x) = -12x + 15 is correct.

 

[Kevin_Axion]

Here is the basic power rule, just apply it to each term: f(x)= n^x therefore,
f'x,\frac{df(x)}{dx}= (n)x^{n-1}.

It's the same thing as: f'(a)=\lim_{h\to 0}{f(a+h)-f(a)\over h}

 

[Mark44]

Here is the basic power rule, just apply it to each term: f(x)= n^x

This is not the power rule. The function you show is an exponential function, not a power function.

A power function is a function of the form xⁿ.

If f(x) = xⁿ, then f'(x) = nx^n-1

therefore,
f'x,\frac{df(x)}{dx}= (x)n^{x-1}.

It's the same thing as: f'(a)=\lim_{h\to 0}{f(a+h)-f(a)\over h}

No, it's not the same thing. You can use the definition of the derivative to get the formula for the derivative of a power function.

 

[Kevin_Axion]

Sorry I messed up the n and x, I fixed it. I know that f'(a)=\lim_{h\to 0}{f(a+h)-f(a)\over h} is essentially the plugging in of points and finding the slope at that point. What I'm essentially saying is that the power rule gives a function in which you can find the slope at a point x and that is what the limit formula does it finds the slope at a point, a. The process is different but the end result is the same, f'x,\frac{df(x)}{dx}= (n)x^{n-1} finds the slope at a, like the limit, if you plug in a. Sorry for my mis-phrasing, I know what you mean though.

 

[Mark44]

You still have n^x in another part, if you want to fix that, too.