Find f′(3) for the function f(x)=√(1+5x), using the definition of derivative.?

Question

I need:
(a) the expression
(b) value of limit

Answer 1

√(1 + 5x + 5h) - √(1+5x) / h



Multiply by conjugate to get:

(1 + 5x + 5h) - (1 + 5x) / (h)(√(1+5x + 5h) + √(1+5x))



1 and 5x will cancel in numerator:



5h / (h)(√(1+5x + 5h) + √(1+5x))



Cancel an h:

5 / (√(1+5x + 5h) + √(1+5x))



Take limit as h-->0

5 / √(1+5x) + √(1+5x)



5 / 2√(1+5x)



That is the derivative.

Now lets plug in 3 to find the derivative evaluated at 3.

(5 / 2√(1 + 15))

5 / 2√16



5/8

Answer 2

Look back at f(x) equal to square root of 1+5x =3 o.o that's mean

f(3)=square root1+15 = square root 16 f(3)=4



value of limit another word domain !!!!

square root function restriction no negative square root .

negative square root mean imaginary number which is pain in the butt to do , also when u graph it ,there aren't any x intercept ... so as long as u don't negative square root it works .

set square root 1+5x =0

then ^2 both side = 1+5x=0 then 5x=-1 x=-1/5 so! x most greater or equal to -1/5

To write it out bracket -1/5,infinity) :D

Answer 3

simply by fact the question shows, this must be solved utilising the definition of a by-product, replace values into the definition of a by-product: f'(x) = lim (h-0) [f(x+h) - f(x)]/h f'(x) = lim (h-0) [a million/[2(x+h)+3] - a million/(2x+3)]/h Evualate f'(x) = lim (h-0) [a million/(2x+2h+3) - a million/(2x+3)]/h come across a user-friendly denominator, (2x+3)(2x+2h+3) f'(x) = lim (h-0) [(2x+3)/(2x+2h+3(2x+3)) - (2x+2h+3)/(2x+3(2x+2h+3)]/h considering that we've user-friendly denominators, combine like words f'(x) = lim (h-0) -2h/( 2x+2h+3(2x+3)) * a million/h pass cancel "h" f'(x) = lim (h-0) -2/(2x+2h+3(2x+3)) ultimately evulate the shrink f'(x) = -2/[(2x+3)(2x+3)] you would be able to examine your answer via looking the by-product utilising the quotient rule. f'(x) = [(2x+3)*0 - a million*(2)]/((2x+3)^2) Simplify to: f'(x) = -2/((2x+3)^2) that's same to our unique answer of: f'(x) = -2/[(2x+3)(2x+3)]