Respuesta :
Answer:
70 is answer
Step-by-step explanation:
Given that a function in x is
[tex]f(x) = 5x^2[/tex]
we have to find f'(7)
we know by derivative rule derivative of a function is
[tex]f'(x) = lim_({h-->0}) \frac{f(x+h)-f(x)}{h}[/tex]
For finding out at 7 we replace x by 7
[tex]f'(7) = lim_({h-->0}) \frac{f(7+h)-f(7)}{h}[/tex]
=[tex]lim\frac{5(7+h)^2-5*7^2}{h} \\= lim \frac{10h*7+h^2}{h} \\= 70+h = 70[/tex]
So f'(7) = 70
answer is 70
Answer:
f'(7)=70
Step-by-step explanation:
We have the definition of the derivative as:
[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]
Now we have a function [tex]f(x)=5x^2[/tex] and we want to approximate the first derivative around x=7, that is [tex]f'(7)[/tex].
We can replace this in the first formula as:
[tex]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[/tex]
Then, the value for f'(7) is:
[tex]f'(7)=10\cdot 7=70[/tex]
