Find the derivative of the following using the definition of derivative. Write the solution of this answer. Use Limits of Delta X approaching zero

Question

3. y = √x

Answer: 1/2√x

4. f(x) = 1/3x

Answer: -1/3x²

Answer 1

Kevin, since you are asking about f(x) = 1/(3x) using definition of derivative, here you go.

f'(x) = lim_h->0 {1/[3(x+h)] - 1/(3x)} / h

This is a complex fraction and is best simplified by multiplying numerator and denominator of the "big" fraction by the LCD of all the "little" denominators. The LCD is 3x(x + h).

If you multiply the 1st term in the numerator by that LCD, the 3 and (x + h) factors cancel and you are left with x. Multiplying the 2nd term of the numerator by the LCD results in the 3 and x canceling and you are left the the (x + h) factor. Taking the LCD times the h in the denominator of the big fraction results in:

h (3x) (x + h).

So here is what we have:

f'(x) = lim_h->0[x - (x + h)]/ [h (3)(x)(x+h)]

The numerator becomes -h and the denominator can be rewritten as 3hx(x+h). As h->0 the factor of h in the numerator and denominator cancel.

f'(x) = lim_h->0 -1/[3x(x+h)] = -1/3x²

Answer 2

I'll do the first one to show the process.

y = √x

y' = lim Δx→0 (√(x+Δx) - √x)/Δx

y' = lim Δx→0 ((√(x+Δx) - √x)/Δx)/Δx)((√(x+Δx)+√x)/(√(x+Δx)+√x))

Remember (a-b)(a+b)= a²-b²

y' = lim Δx→0 (x+Δx-x)/(Δx(√(x+Δx)+√x))

y' = lim Δx→0 Δx/(Δx(√(x+Δx)+√x)) Δx's cancel

y' = lim Δx→0 1/(√(x+Δx)+√x)

Take the limit by setting Δx to 0

y' = 1/(√x + √x) = 1/(2√x)

Answer 3

This is the trick you need for the square root derivative: