Question
Find the derivative of the function using the definition of a derivative.
F(x)=squareroot(1-3x)
Please show work for understanding, thank you! :)
F(x)=squareroot(1-3x)
Please show work for understanding, thank you! :)
Answers
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)
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)
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