Find the derivative of 2x+3/ 3x-2
using first principle method.
To find the derivative of a function using the first principle method, we need to use the definition of a derivative. The definition of a derivative states that the derivative of a function f(x) at a point x is given by the following limit:
f'(x) = lim(h→0) [f(x + h) - f(x)] / h
In this case, we need to find the derivative of the function f(x) = (2x + 3) / (3x - 2). Let's use the first principle method to evaluate it.
Step 1: Start with the function f(x) = (2x + 3) / (3x - 2).
Step 2: Expand the numerator using the distributive property:
f(x) = 2x/ (3x - 2) + 3 / (3x - 2).
Step 3: Apply the definition of the derivative:
f'(x) = lim(h→0) [f(x + h) - f(x)] / h.
Step 4: Substitute the value of f(x + h) - f(x) into the definition of the derivative:
f'(x) = lim(h→0) [(2(x + h) + 3) / (3(x + h) - 2) - (2x + 3) / (3x - 2)] / h.
Step 5: Simplify the expression inside the limit:
f'(x) = lim(h→0) [(2x + 2h + 3) / (3x + 3h - 2) - (2x + 3) / (3x - 2)] / h.
Step 6: Combine the two fractions by finding a common denominator:
f'(x) = lim(h→0) [(2x + 2h + 3)(3x - 2) - (2x + 3)(3x + 3h - 2)] / [(3x + 3h - 2)(3x - 2)] / h.
Step 7: Expand and simplify the expression further:
f'(x) = lim(h→0) [(6x^2 + 4hx - 4x - 2h + 9x - 6 - 6x^2 - 6hx + 4x + 6) / [(3x + 3h - 2)(3x - 2)] / h.
Step 8: Simplify the expression inside the limit:
f'(x) = lim(h→0) [(4hx - 2h + 9x - 6 - 6hx + 4x + 6) / [(3x + 3h - 2)(3x - 2)] / h.
Step 9: Combine like terms:
f'(x) = lim(h→0) [(4hx - 6hx) + (9x + 4x - 2h - 6 + 6)] / [(3x + 3h - 2)(3x - 2)] / h.
Step 10: Simplify the expression further:
f'(x) = lim(h→0) [-2hx + 13x] / [(3x + 3h - 2)(3x - 2)] / h.
Step 11: Factor out an h from the numerator:
f'(x) = lim(h→0) h(-2x + 13x) / [(3x + 3h - 2)(3x - 2)] / h.
Step 12: Cancel out h from the numerator and denominator:
f'(x) = lim(h→0) -2x + 13x / (3x + 3h - 2)(3x - 2).
Step 13: Take the limit as h approaches 0:
f'(x) = -2x + 13x / (3x - 2)^2.
Step 14: Simplify the expression:
f'(x) = (11x) / (3x - 2)^2.
Therefore, the derivative of the function f(x) = (2x + 3) / (3x - 2) using the first principle method is f'(x) = (11x) / (3x - 2)^2.
ou meant:
f(x) = y = (2x+3)/(3x-2)
so f(x+h) = (2x + 2h + 3)/(3x + 3h - 2)
dy/dx = lim [ (2x+2h+3)/(3x+3h-2) - (2x+3)/(3x-2) ]/h , as h --->0
= lim [ ((2x + 2h + 3)(3x-2) - (2x+3)(3x+3h+2))/((3x+3h-2)(3x+h))]/h
= lim [ (5h/((3x+3h-2)(3x-2))]/h , as h -->0
= lim 5/((3x+3h-2)(3x-2)) , as h -->0
= 5/(3x-2)^2
The answer is= -13/(3x-2)^2. But the answer they are post is incorrect answer.
you meant:
f(x) = y = (2x+3)/(3x-2)
so f(x+h) = (2x + 2h + 3)/(3x + 3h - 2)
dy/dx = lim [ (2x+2h+3)/(3x+3h-2) - (2x+3)/(3x-2) ]/h , as h --->0
= lim [ ((2x + 2h + 3)(3x-2) - (2x+3)(3x+3h+2))/((3x+3h-2)(3x+h))]/h
= lim [ (5h/((3x+3h-2)(3x-2))]/h , as h -->0
= lim 5/((3x+3h-2)(3x-2)) , as h -->0
= 5/(3x-2)^2