Question

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.)

Answer 1

f'(7)=70

Explanation:

We have the definition of the derivative as:

`f'(x)= \lim_(h \to 0) (f(x+h)-f(x))/(h)`

Now we have a function
`f(x)=5x^2` and we want to approximate the first derivative around x=7, that is
`f'(7)`.

We can replace this in the first formula as:

`f'(x)= \lim_(h \to 0) (f(x+h)-f(x))/(h)= \lim_(h \to 0) (5(x+h)^2-5x^2)/(h)\\\\f'(x)=\lim_(h \to 0) (5(x^2+2xh+h^2-x^2))/(h)\\\\f'(x)=\lim_(h \to 0)(5(2xh+h^2))/(h)\\\\f'(x)=\lim_(h \to 0)5(2x+h)\\\\f'(x)=10x+lim_(h \to 0)h=10x+0=10x`

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

`f'(7)=10\cdot 7=70`

Answer 2

70 is answer

Explanation:

Given that a function in x is

`f(x) = 5x^2`

we have to find f'(7)

we know by derivative rule derivative of a function is

`f'(x) = lim_({h-->0}) (f(x+h)-f(x))/(h)`

For finding out at 7 we replace x by 7

`f'(7) = lim_({h-->0}) (f(7+h)-f(7))/(h)`

=
`lim(5(7+h)^2-5*7^2)/(h) \\= lim (10h*7+h^2)/(h) \\= 70+h = 70`

So f'(7) = 70

answer is 70