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

Answer:

f'(7)=70

Step-by-step explanation:

We have the definition of the derivative as:

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

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:

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}= \lim_{h \to 0} \dfrac{5(x+h)^2-5x^2}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{5(x^2+2xh+h^2-x^2)}{h}\\\\f'(x)=\lim_{h \to 0}\dfrac{5(2xh+h^2)}{h}\\\\f'(x)=\lim_{h \to 0}5(2x+h)\\\\f'(x)=10x+lim_{h \to 0}h=10x+0=10x$

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

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

Answer 2

Answer:

70 is answer

Step-by-step 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

$f'(x) = lim_({h-->0}) \frac{f(x+h)-f(x)}{h}$

For finding out at 7 we replace x by 7

$f'(7) = lim_({h-->0}) \frac{f(7+h)-f(7)}{h}$

=$lim\frac{5(7+h)^2-5*7^2}{h} \\= lim \frac{10h*7+h^2}{h} \\= 70+h = 70$

So f'(7) = 70

answer is 70