Using the definition of derivative find f'(x) when f(x) = (1/sqr x) + x.

Answer 1

f(x+h) = 1/√(x+h) + x+h
f(x) = 1/√x + x

f(x+h)-f(x) = 1/√(x+h) - 1/√x + h
= (x+h)^(-1/2) - x^(-1/2) + h

From the binomial theorem,
(a+b)^n = a^n + n a^n-1 b + ...

(x+h)^(-1/2) = x^(-1/2) - 1/2 x^(-3/2)h + 3/8 x^(-5/2)h^2 + ...
where ... is stuff with higher powers of h.

Now, we just subtract
f(x+h)-f(x) = h -1/2 x^(-3/2)h + 3/8 x^(-5/2)h^2 + ...

Now for the quotient

(f(x+h)-f(x))/h = 1 - 1/2 x^-3/2 + 3/8 x^(-5/2)h + ...

Now take the limit as h->0 and it all vanishes except

f'(x) = 1 - 1/2 x^-3/2

Answer 2

sometimes called:
find the derivative from First Priciples

f(x) = 1/√x + x
f(x+h) = 1/√(x+h) + x+h

f'(x) = Lim ( f(x+h) - f(x) )/h , as h --->0
= lim( 1/√(x+h) + x+h - (1/√x + x )/h as h--->0
= lim ( (1/√(x+h) - 1/√x)/h + h/h) as h--> 0
= lim ( (1/√(x+h) - 1/√x)/h* (√x√(x+h)/(√x√x+h) + h/h) as h--> 0
= Lim ( (√x - √(x+h)/(h√x√(x+h) ) + h/h) as h --->0
= Lim ( (√x - √(x+h)/(h√x√(x+h)*(√x + √(x+h)/(√x + √(x+h)) ) + h/h) as h --->0
= lim (x - x - h)/(h√x√(x+h)(√x + √(x+h)) + h/h
= lim -1/(√x√(x+h)(√x + √(x+h) ) + 1 , as h --->0
= -1/(√x√x(√x + √x) + 1
= -1/(x(2√x)) + 1
or -1/(2(√x)^3)

Wheewww!

Answer 3

Darn!!! last line, forgot the +1 , but you figured that out, didn't you?