trig limits\n$lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1;lim_{\theta\rightarrow0}\frac{cos\theta…

trig limits\n$lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1;lim_{\theta\rightarrow0}\frac{cos\theta - 1}{\theta}=0$\nlinearization of $f$ at $a$\n$l(x)=f(a)+f(a)(x - a)$\nquestion 8\nfind $f(x)$ for $f(x)=\tan x$.\n$f(x)=-csc^{2}x$\n$f(x)=sec^{2}x$\nno correct answer choice is given.\n$f(x)=sec x\tan x$\n$f(x)=-csc xcot x$

trig limits\n$lim_{\theta\rightarrow0}\frac{sin\theta}{\theta}=1;lim_{\theta\rightarrow0}\frac{cos\theta - 1}{\theta}=0$\nlinearization of $f$ at $a$\n$l(x)=f(a)+f(a)(x - a)$\nquestion 8\nfind $f(x)$ for $f(x)=\tan x$.\n$f(x)=-csc^{2}x$\n$f(x)=sec^{2}x$\nno correct answer choice is given.\n$f(x)=sec x\tan x$\n$f(x)=-csc xcot x$

Answer

Explanation:

Step1: Recall the derivative formula

The derivative of $\tan x$ can be derived from the quotient - rule. Since $\tan x=\frac{\sin x}{\cos x}$, and by the quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = \sin x$, $u'=\cos x$, $v=\cos x$, $v'=-\sin x$.

Step2: Apply the quotient - rule

$f'(x)=\frac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^{2}x}=\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}$.

Step3: Use the trigonometric identity

Since $\sin^{2}x+\cos^{2}x = 1$, then $f'(x)=\frac{1}{\cos^{2}x}=\sec^{2}x$.

Answer:

B. $f'(x)=\sec^{2}x$