Can someone please explain this to me using simple words
#Derivitive
glass junco
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Can someone please explain this to me using simple words
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dusky lance
Sure. The definition of the derivative is:
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
$$
So when we take f(x) := kg(x) you have:
$$
\begin{aligned*}
f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\
&= \lim_{h \to 0} \frac{kg(x+h)-kg(x)}{h}\
&= k\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}
\end{aligned*}
$$
In fact, when we differentiate, we are trying to approximate the function by a line.
So, if you "enlarge" your values (or reduce them if $|k| \leq 1$) you will also change this line by the scalar $k$.
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Heolorus
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dusky lance
I'm not able to find the error
tepid lodge
I'm not able to find the error
You can't use both $$ and begin align
If you are using $$ then you have to type aligned* instead of align*
Sure. The definition of the derivative is:
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
$$
So when we take $f(x) := kg(x)$ you have:
\begin{align*}
f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\
&= \lim_{h \to 0} \frac{kg(x+h)-kg(x)}{h}\
&= k\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}
\end{align*}
In fact, when we differentiate, we are trying to approximate the function by a line.
So, if you "enlarge" your values (or reduce them if $|k| \leq 1$) you will also change this line by the scalar $k$.
fallow lavaBOT
Vee
dusky lance
why is it in white ?
glass junco
ahh i see