Limit form of derivatives help?

Question

Alright, I'm having trouble with these two problems. I'm supposed to use the limit form of the derivative to find dy/dx. Simple enough for all of the other problems except these two:

y = √x

and

y = 1 / (x^2)

Any help with these would be greatly appreciated.

Answer 1

d(√x)/dx
[h→0]lim (√(x+h) - √x)/h

Multiply both numerator and denominator by √(x+h) + √x:

[h→0]lim (x+h - x)/(h(√(x+h) + √x))
[h→0]lim h/(h(√(x+h) + √x))
[h→0]lim 1/(√(x+h) + √x)
1/(√(x+0) + √x)
1/(√x + √x)
1/(2√x)

---------------

d(1/x²)/dx
[h→0]lim (1/(x+h)² - 1/x²)/h
[h→0]lim (x²/(x²(x+h)²) - (x+h)²/(x²(x+h)²))/h
[h→0]lim ((x² - (x+h)²)/(x²(x+h)²))/h
[h→0]lim (x² - (x+h)²)/(x²h(x+h)²)
[h→0]lim (x² - x² - 2xh + h²)/(x²h(x+h)²)
[h→0]lim (-2xh + h²)/(x²h(x+h)²)
[h→0]lim (-2x + h)/(x²(x+h)²)
(-2x + 0)/(x²(x+0)²)
-2x/x⁴
-2/x³

And we are done.

Answer 2

Set y(x) = √x. The derivative of y(x) at a point a>0 is
by definition the limit

y'(a) = lim_{x->a} (√x - √a)/(x-a)

Write x-a = (√x - √a) (√x + √a) (valid for x,a>0) and replace
it in the previous expression, then

y'(a) = lim_{x->a} 1/ (√x + √a) = 1/(2 √a)

by continuity of the square root.


For y(x) = y = 1 / (x^2). Take a different from 0:

y'(a) = lim_{x->a} (1/x^2- 1/a^2)/(x-a)

The term in the limit can be simplified:

(1/x^2- 1/a^2)/(x-a)= (a^2 -x^2)/(a^2 x^2 (x-a))
= -(a+x)/(a^2 x^2 )

so that

y'(a) = lim_{x->a} -(a+x)/(a^2 x^2 ) = - 2/a^3