Using the limit definition, how do you find the derivative of f(x)= 2x^2-x?

1 Answer
Feb 11, 2016

f'(x)=4x-1

Explanation:

By limit definition, the derivative of any function is
f'(x)=lim_(h\to0) (f(x+h)-f(x))/h

Now, for this particular question, we have f(x)=2x^2-x
So, using the derivative definition, we have
f'(x)=lim_(h\to0)((2(x+h)^2-(x+h))-(2x^2-x))/h

Rearranging the equation and separating to have the x parameters of same power, we have
f'(x)=lim_(h\to0)((2(x+h)^2-2x^2)/h-(x+h-x)/h)

In the right side, the second term is simple so we'll directly get it as 1. In the first term, notice that we can take 2 as a common term. Next, by taking a=x+h and b=x, we can use the identity a^2-b^2=(a+b)(a-b) for which we get the answer (2x+h)(h) Substituting that above, we get
f'(x)=lim_(h\to0)(2(2x+h)h/h-1)=lim_(h\to0)(2(2x+h)-1)

Now, since it's not possible to reduce the terms any further, we can finally just take h=0 and get what we wanted. Hence you should get the answer I've written way above.