📝 Derivatives of tan cot sec csc (16 MCQs)

📖 Explanation: The derivative of the tangent function follows directly from the quotient rule or from the identity $\tan x = \sin x / \cos x$. Differentiating gives $\frac{d}{dx}\tan x = \sec^2 x$. This result is standard and appears in most calculus tables, making option A the correct choice.

📖 Explanation: Since $\frac{d}{dx}\tan x = \sec^2 x$ and $\sec x = 1/\cos x$ is positive for $0<x<\pi/2$, the square $\sec^2 x$ is always positive. Therefore the tangent curve is increasing throughout the interval, giving a positive slope.

📖 Explanation: At $x=\pi/4$, $\sec^2(\pi/4) = (\sqrt{2})^2 = 2$. Doubling this rate yields $2 \times 2 = 4$. Hence the derivative after doubling would be 4, which corresponds to option C.

📖 Explanation: Using the chain rule, (f\circ g)&#039;(0)=\sec^2(g(0))\cdot a. If we choose $b=0$, then $\sec^2(0)=1$ and the equation becomes $a=9$. Thus $a=9$ satisfies the condition, giving option C.

📖 Explanation: The instantaneous velocity is the derivative s&#039;(t)=\sec^2 t, which is never zero because a square of a real number is always non‑negative and equals zero only when the base is zero, which never occurs for $\sec t$. Hence no interval contains a zero velocity.

📖 Explanation: When $\cos x = 1/2$, $\sec x = 2$ and $\tan x = \sqrt{3}$. The derivative of $\tan x$ is $\sec^2 x = 4$. The derivative of $\sec x$ is $\sec x \tan x = 2\sqrt{3}\approx3.46$. Hence the derivative of tan is larger, confirming option A.

📖 Explanation: The definition of derivative at a point $a$ is $\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$. Substituting $f(x)=\tan x$ and $a=0$ yields the first limit, making it the correct representation.

📖 Explanation: Differentiability requires matching left‑hand and right‑hand derivatives at $x=\pi/4$. The left derivative is $\sec^2(\pi/4)=2$. The right derivative equals the constant slope $k$. Setting $k=2$ ensures continuity of the derivative, so option B is correct.

📖 Explanation: The finite‑difference approximation replaces the limit definition of the derivative with a non‑zero step $h=0.1$. The dominant discrepancy arises from ignoring terms of order $h$ and higher, i.e., truncation error, making option A the primary source of inaccuracy.

📖 Explanation: At $x=\pi/3$, $\cos(\pi/3)=1/2$. Hence $\sec(\pi/3)=2$ and $\sec^2(\pi/3)=4$. The derivative of $\tan x$ equals $\sec^2 x$, so the correct value is 4, which is listed as option B after re‑ordering.

📖 Explanation: Differentiating $V=\tan(kW)$ with respect to $W$ gives $\frac{dV}{dW}=k\sec^2(kW)$. Thus the instantaneous change depends on both the factor $k$ and the squared secant term, corresponding to option C.

📖 Explanation: Applying the chain rule, h&#039;(t)=\sec^2(3t^2+2t)\cdot(6t+2). The inner derivative of $3t^2+2t$ is $6t+2$. Hence the first option correctly represents the derivative.

📖 Explanation: Adding a constant to a function does not affect its derivative because the constant's derivative is zero. Therefore the derivative of the shifted function is identical to the original derivative, leaving the graph unchanged.

📖 Explanation: Differentiating $\tan x = \sin x / \cos x$ and simplifying uses the Pythagorean identity $\tan^2 x + 1 = \sec^2 x$. This identity shows that the derivative simplifies to $\sec^2 x$, making option A the direct link.

📖 Explanation: Vertical asymptotes of $\tan x$ occur where $\cos x = 0$. Near these points, $\sec x = 1/\cos x$ grows without bound, and squaring it makes $\sec^2 x$ tend to infinity. Hence the derivative blows up, corresponding to option B.

📖 Explanation: The Mean Value Theorem states that there exists a point $c$ in $(a,a+\Delta)$ where the instantaneous derivative equals the average rate of change. Since the derivative of $\tan x$ is $\sec^2 x$, we have $\sec^2(c)=3$. Thus option A is guaranteed.