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) = a + √a
(a) Find the derivative of the function using the definition of derivative.
3 answers
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)
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.
1 answer