By determining f prime left parenthesis x right parenthesis equals modifyingbelow lim with h right arrow 0 startfraction f left parenthesis x plus h right parenthesis minus f left parenthesis x right parenthesis over h endfractionf′(x)=limh→0 f(x+h)−f(x) h, find f prime left parenthesis 7 right parenthesisf′(7) for the given function. f left parenthesis x right parenthesis equals 5 x squaredf(x)=5x2 f prime left parenthesis 7 right parenthesisf′(7)equals=nothing (simplify your answer.)

70 is answer

Step-by-step explanation:

Given that a function in x is

we have to find f'(7)

we know by derivative rule derivative of a function is

For finding out at 7 we replace x by 7

=

So f'(7) = 70

answer is 70

f'(7)=70

Step-by-step explanation:

We have the definition of the derivative as:

Now we have a function and we want to approximate the first derivative around x=7, that is .

We can replace this in the first formula as:

Then, the value for f'(7) is:

were is the equation

i wish i could u

step-by-step explanation:

i wish i could u