Simplifying for derivative

Question

consider the function f(x) = 4sqrt(x)-1

a) Simplify as much as possible in f(4+h)-f(4)/h

b) Use the result to find the derivative of f'(4)

Answer 1

f(4+h) = 4 * sqrt( 4+h) - 1

f(4) = 4 * sqrt(4) - 1 = 4 * 2 - 1 = 8 - 1 = 7

f(4+h) - f(4) = 4 * sqrt(4 + h ) - 1 - 7

= 4 * sqrt(4+h) - 8

the difference quotient is then [4 * sqrt(4+h) - 8] / h

= 4 ( sqrt(4+h) - 2) / h

Rationalizing the numerator:

4 ( sqrt(4+h) - 2) ( sqrt(4+h) + 2) / [ h ( sqrt( 4+h)+2) ] =

4 ( (4+h) - 4 ) / [ h ( sqrt( 4+h) + 2 ] =

4 h / [ h ( sqrt( 4+h) + 2 ] =

4/[ sqrt( 4+h) + 2]

As h--->0 the difference quotient tends to 1

Now differentiation for real:

y = 4 * x^(1/2) - 1

y ' = 4 ( 1/2) * x^ ( 1/2)

= 2 / sqrt(x)

f'(4) = 2/sqrt(4) = 2/2 = 1

Answer 2

f(4+h) = 4sqrt(4+h)-1

f(4) = 4sqrt(4)-1

So f(4+h) - f(4) = 4sqrt(4+h)-1 - (4sqrt(4)-1) = 4sqrt(4+h)-1 - 4sqrt(4) + 1 = 4sqrt(4+h) - 8 = 4(sqrt(4+h) - 2)

The derivative is defined as the limit as h→0 of [f(x+h) - f(x)]/h so in this case the derivative = the limit as h→0 of 4(sqrt(4+h) - 2)/h

To resolve this limit, multiply by the conjugate [sqrt(4+h) + 2)]/[sqrt(4+h) + 2)] to get the limit as h→0 of 4(4+h-4)/[h(sqrt(4+h)+2)] = the limit as h→0 of 4h/[h(sqrt(4+h)+2)] = the limit as h→0 of 4/[sqrt(4+h)+2)] = 4/4 = 1

Based on the given expression, we are finding the derivative at x=4 so the derivative at x=4 is 1