Find the derivative of the function using the definition of a derivative.
F(x)=squareroot(1-3x)
Please show work for understanding, thank you! :)
recall that (a+b)^n = a^n + na^(n-1)*b + n(n-1)/2 a^(n-2)*b^2 + ...
So,
(1-3(x+h))^1/2 = ((1-3x) - 3h)^1/2
= (1-3x)^1/2 + (1/2)(1-3x^(-1/2)*(-3h) + (*)(-3h)^2 + ...
f(x+h) - f(x) =
(1-3x)^1/2 + (1/2)(1-3x^(-1/2)*(-3h) + (*)(-3h)^2 + ... - (1-3x)^1/2
= (1/2)(1-3x^(-1/2)*(-3h) + (*)(-3h)^2 + ...
now divide that by h to get
(1/2)(1-3x^(-1/2)*(-3) + (*)(9h) + ...
Now take the limit as h->0.
The (9h) term and all the other higher-power-of-h terms vanish, leaving
-3/2 (1-3x)^(-1/2)
To find the derivative of the function F(x) = √(1 - 3x) using the definition of a derivative, we need to start by applying the limit definition of the derivative:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Let's proceed with the steps:
1. Replace f(x) with the given function, F(x):
f'(x) = lim(h->0) [F(x + h) - F(x)] / h
2. Substitute F(x) with its expression:
f'(x) = lim(h->0) [√(1 - 3(x + h)) - √(1 - 3x)] / h
3. Simplify by removing the square roots in the numerator using the difference of squares:
f'(x) = lim(h->0) [(1 - 3(x + h)) - (1 - 3x)] / h / (√(1 - 3(x + h)) + √(1 - 3x))
4. Expand the expression:
f'(x) = lim(h->0) [1 - 3x - 3h - 1 + 3x] / h / (√(1 - 3(x + h)) + √(1 - 3x))
5. Cancel out terms that simplify to 0:
f'(x) = lim(h->0) [-3h] / h / (√(1 - 3(x + h)) + √(1 - 3x))
6. Simplify further by canceling out the "h" in the numerator and denominator:
f'(x) = lim(h->0) -3 / (√(1 - 3(x + h)) + √(1 - 3x))
7. Take the limit as h approaches 0:
f'(x) = -3 / (√(1 - 3x) + √(1 - 3x))
8. Simplify the expression further:
f'(x) = -3 / (2√(1 - 3x))
Therefore, the derivative of the function F(x) = √(1 - 3x) using the definition of a derivative is f'(x) = -3 / (2√(1 - 3x)).