Question

Demonstre, pela definição de derivada, que se f(x) = senx, então f ′ (x) = cox.

Answer 1

Boa noite Lanny!

Solução!

$f(x)=senx\\\\\ f'(x)=cosx$


$f'(x)= \displaystyle\lim_{t \to 0} \dfrac{f(x0+t)-f(x0)}{t}\\\\\\\\ f'(x)= \displaystyle\lim_{t \to 0} \dfrac{sen(x0+t)-sen(x0)}{t}\\\\\\\\ Lembrando~~Sen(a+b)\\\\\\\\ f'(x)= \displaystyle\lim_{t \to 0} \dfrac{sen(x0).cost+sent.cos(x0)sen(x0)}{t}$



$f'(x)= \displaystyle\lim_{t \to 0} \frac{sen(x0)[cost-1]+sent.cos(x0)}{t}\\\\\\\\ Lembrando~~que~~(a-b).(a+b)=(a^{2}-b^{2})\\\\\\\\\\\ f'(x)= \displaystyle\lim_{t \to 0} \frac{sen(x0)[cost-1].[cost+1]+sent.cos(x0)}{t[cost+1} \\\\\\\\ f'(x)= \displaystyle\lim_{t \to 0} \frac{sen(x0)[Cos^{2}-1 ]+sent.cos(x0)}{t[cost+1} \\\\\\\\ Relac\~ao ~~fundamental~~da~~trigonometria!\\\\\\ Sen^{2}x+Cos^{2}=1 \Rightarrow Cos^{2}-1=-Sen^{2}x$



$f'(x)= \displaystyle\lim_{t \to 0} \frac{sen(x0)[-Sen^{2}]}{t[cost+1]}+ \frac{sent.cos(x0)}{t} \\\\\\\\ f'(x)= \displaystyle\lim_{t \to 0} \frac{sen(x0)[-Sen^{2}x}{t[cost+1}]+ \frac{sent.cos(x0)}{t} \\\\\\\ f'(x)= \displaystyle\lim_{t \to 0} \frac{-sen(x0)Senx.Senx}{t[cost+1]}+ \frac{sent.cos(x0)}{t} \\\\\\\ Lembrando~~do~~limite~~fundamental~~trigonometrico!\\\\\\\\\ f'(x)= \displaystyle\lim_{t \to 0} 0+ 1.cos(x0) \\\\\\\\\\\ f'(x)= \displaystyle\lim_{t \to 0} cos(x0)$



$Esta ~~provado!\\\\\\\ \boxed{f'(x)=cosx}$

Boa noite!
Bons estudos!