Use the definition principle to find the derivative of the following function?

Question

f(x) = 1 / (2x+3)

Answer 1

d(f(x))/dx = -2/(2x+3)^2

you can write f(u) = d(u^-1)/du where u =2x+3

the derivative of u(x)^n is u(x)^(n-1) * du/dx

So for u^n-1 derivative = u^(-1-1) =u^(-2) = 1/u^2
and derivative of (2x+3) =2
so , you find the result in first line

Answer 2

As the question suggests, this must be solved using the definition of a derivative, substitute values into the definition of a derivative:
f'(x) = lim (h-0) [f(x+h) - f(x)]/h

f'(x) = lim (h-0) [1/[2(x+h)+3] - 1/(2x+3)]/h

Evualate
f'(x) = lim (h-0) [1/(2x+2h+3) - 1/(2x+3)]/h

Find a common denominator, (2x+3)(2x+2h+3)
f'(x) = lim (h-0) [(2x+3)/(2x+2h+3(2x+3)) -
(2x+2h+3)/(2x+3(2x+2h+3)]/h

Since we have common denominators, combine like terms
f'(x) = lim (h-0) -2h/( 2x+2h+3(2x+3)) * 1/h

Cross cancel "h"
f'(x) = lim (h-0) -2/(2x+2h+3(2x+3))

Finally evulate the limit
f'(x) = -2/[(2x+3)(2x+3)]


You may check your answer by finding the derivative using the quotient rule.

f'(x) = [(2x+3)*0 - 1*(2)]/((2x+3)^2)

Simplify to:
f'(x) = -2/((2x+3)^2)

This is identical to our original answer of:
f'(x) = -2/[(2x+3)(2x+3)]

Answer 3

sooo.... ill do it the same way hokay
f(x)=(2x-3)^-1
so f'(x)= -2((2x-3)^-2)

.... if u dont follow... pay attention in class... but u do the derivative of the inside and outside but the inside remains the same...