Find the derivative by definition of
F(x)= x/x+1
by first principles:
f(x) = x/(x+1), notice the necessary brackets
f(x+h) = (x+h)/(x+h + 1)
F'(x) = limit ((x+h)/(x+h + 1) - x/(x+1) )/(x+h - x), as h ----> 0
= lim [( (x+h)(x+1) - x(x+h + 1) ) / ((x+1)(x+h+1) ]/h , as h ---> 0
= lim [ (x^2 + hx + x + h - x^2 - xh - x)/((x+1)(x+h+1) ] / h , as h ---> 0
= lim [ h/((x+1)(x+h+1)]/h , as h ---> 0
= 1/((x+1)(x+h+1) , as h ---> 0
= 1/(x+1)^2
Thank you!
To find the derivative of F(x) = x / (x + 1) using the definition of a derivative, we need to apply the limit definition of the derivative, which is:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Let's begin by applying this definition to our function F(x).
f'(x) = lim(h->0) [F(x + h) - F(x)] / h
Evaluate F(x + h):
F(x + h) = (x + h) / ((x + h) + 1) = (x + h) / (x + h + 1)
Now substitute F(x + h) and F(x) back into the definition:
f'(x) = lim(h->0) [(x + h) / (x + h + 1) - x / (x + 1)] / h
Combine the fractions:
f'(x) = lim(h->0) [(x + h)(x + 1) - x(x + h + 1)] / (h(x + h + 1))
Expand the terms:
f'(x) = lim(h->0) [x^2 + h + x + hx + x + h^2 + h - x^2 - hx - x] / (h(x + h + 1))
Simplify:
f'(x) = lim(h->0) [h^2 + 2h] / (h(x + h + 1))
Cancel out the common factors:
f'(x) = lim(h->0) [h(h + 2)] / (h(x + h + 1))
Now, we can cancel out the h in the numerator and denominator:
f'(x) = lim(h->0) (h + 2) / (x + h + 1)
As h approaches 0, we can substitute it into the equation:
f'(x) = (0 + 2) / (x + 0 + 1)
Simplify:
f'(x) = 2 / (x + 1)
Therefore, the derivative of F(x) = x / (x + 1) using the definition of a derivative is f'(x) = 2 / (x + 1).