f(a) = a + √a
(a) Find the derivative of the function using the definition of derivative.
The derivative f'(a) is the sum of the derivative of a and the derivative of sqrt a. In this case, a is treated as a variable, not a constant.
The answer is 1 + 1/(sqrt a)
f(a + h) = a + h + (a+h)^1/2
f(a) = a + a^(1/2)
f(a+h) - f(a) = h + (a+h)^1/2 - a^1/2
binomial series for q<p: (p+q)^n= p^n +n p^(n-1) q + n(n-1)/2! p^(n-2)q^2 ....
so
for small h
f(a+h)-f(a) = h + a^1/2 + (1/2) a^(-1/2)h + (1/2)(-1/2)/2*a^(-1.5)h^2.. -a^1/2)
f(a+h)-f(a) = h + (1/2)a^(-1/2) h - 1/8 a^-1.5 h^2 ....
divide by h
(f(a+h)-f(a))/h = 1 + (1/2) a^-1/2 - (1/8) a^-1.5) h ...
let h-->0
df/da---> 1 + (1/2)a^(-1/2)
whoops, I forgot the (1/2). Damon also provided the derivation you wanted.
To find the derivative of a function using the definition of derivative, we need to apply the limit definition of the derivative equation.
Let's start by writing down the definition of the derivative:
f'(a) = lim(h → 0) [f(a + h) - f(a)] / h
Now, let's substitute the given function f(a) = a + √a into the definition of the derivative:
f'(a) = lim(h → 0) [(a + h + √(a + h)) - (a + √a)] / h
Next, let's simplify the expression inside the limit by expanding the terms and canceling out the common terms:
f'(a) = lim(h → 0) [a + h + √(a + h) - a - √a] / h
f'(a) = lim(h → 0) [h + √(a + h) - √a] / h
Now, let's try to simplify this expression further. We can rationalize the numerator by multiplying by the conjugate of the expression √(a + h) - √a:
f'(a) = lim(h → 0) [(h + √(a + h) - √a) * (√(a + h) + √a)] / (h * (√(a + h) + √a))
Expanding the numerator using the FOIL method:
f'(a) = lim(h → 0) [h√(a + h) + h√a + h(a + h) - a - √a√(a + h) - √a√a] / (h * (√(a + h) + √a))
Simplifying:
f'(a) = lim(h → 0) [h√(a + h) + h√a + h(a + h) - a - √(a(a + h)) - √(a^2)] / (h * (√(a + h) + √a))
Canceling out the common terms in the numerator and denominator:
f'(a) = lim(h → 0) [h√(a + h) + h√a + h(a + h) - a - √(a(a + h)) - √(a^2)] / (h * (√(a + h) + √a))
Now, we can further simplify by dividing each term in the numerator by h:
f'(a) = lim(h → 0) [√(a + h) + √a + (a + h) - a - (√(a(a + h)) + √(a^2))] / (√(a + h) + √a))
We can now take the limit as h approaches 0:
f'(a) = [√(a + 0) + √a + (a + 0) - a - (√(a(a + 0)) + √(a^2))] / (√(a + 0) + √a))
f'(a) = [√a + √a + a - a - (√(a * 0) + √(a^2))] / (√a + √a))
Simplifying further:
f'(a) = [2√a] / [2√a]
Finally, after canceling out the common terms, we get:
f'(a) = 1
Therefore, the derivative of the function f(a) = a + √a using the definition of derivative is simply 1.