question 10 (1 point)\nthe antiderivative of $f(x)$ which is denoted as $int f(x)dx$ is the function $f(x)$…

question 10 (1 point)\nthe antiderivative of $f(x)$ which is denoted as $int f(x)dx$ is the function $f(x)$ whose derivative is $f(x)$. which functions could be the antiderivative of $intsec^{2}dx$?\n$f(x)=cot x$\n$f(x)=8 + \tan x$\n$f(x)=\tan x$\n$f(x)=-cot x$\n$f(x)=8 - \tan x$

question 10 (1 point)\nthe antiderivative of $f(x)$ which is denoted as $int f(x)dx$ is the function $f(x)$ whose derivative is $f(x)$. which functions could be the antiderivative of $intsec^{2}dx$?\n$f(x)=cot x$\n$f(x)=8 + \tan x$\n$f(x)=\tan x$\n$f(x)=-cot x$\n$f(x)=8 - \tan x$

Answer

Explanation:

Step1: Recall derivative formulas

We know that the derivative of $\tan x$ with respect to $x$ is $\sec^{2}x$, i.e., $\frac{d}{dx}(\tan x)=\sec^{2}x$. Also, the derivative of a constant $C$ is $0$, i.e., $\frac{d}{dx}(C) = 0$.

Step2: Analyze antiderivative

The antiderivative of $\sec^{2}x$ is a function $F(x)$ such that $F'(x)=\sec^{2}x$. Since $\frac{d}{dx}(\tan x)=\sec^{2}x$ and $\frac{d}{dx}(C + \tan x)=\frac{d}{dx}(C)+\frac{d}{dx}(\tan x)=0+\sec^{2}x=\sec^{2}x$ for any constant $C$.

Answer:

B. $F(x)=8 + \tan x$, C. $F(x)=\tan x$