SOLUTION: Find the derivative of ƒ(x)=x*2/3-9x from first principle.

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Question 1119494: Find the derivative of ƒ(x)=x*2/3-9x from first principle.
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
Is it really +%282%2F3%29x+-9x%29+ as you have posted?

My guess is you intended +2x+%2F+%283-9x%29+ ??

You should use parentheses, otherwise we are left guessing. Many tutors will ignore questions that require guessing.

I am guessing +f%28x%29+=+2x+%2F+%283-9x%29+
f'(x) = Lim(h—>0) { ( f(x+h)-f(x) ) / h } —> in words: "the limit of a change in the function value, divided by a small change in x, as that small change in x goes to zero."

f'(x) = Lim(h—>0)

After some algebra (lots of things will cancel) ….

f'(x) = Lim(h—>0)

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Check: using the quotient rule for derivatives (f'(n/d) = (d*n' - n*d') / n^2 ):

f'(x) = +%28%283-9x%29%282%29+-+%282x%29%28-9%29%29+%2F+%283-9x%29%5E2++
= ++6%2F%289%2A%289x%5E2-6x%2B1%29%29+
= +++2%2F%283%283x-1%29%5E2%29+ (ok)

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It looks as though you are studying Calculus. If so, you should study this concept and understand it fully because it is truly fundamental to half of the entire subject. It is not very complex. You may wish to start with a simple function such as f(x) = x^2 and draw out f(x), f(x+h), x, x+h on graph paper so it makes the concept sink in.