📝 Higher order derivatives second third derivative (16 MCQs)

📖 Explanation: The fourth derivative can be written either as the superscript notation $f^{(4)}(x)$ or using the differential operator $\frac{d^{4}}{dx^{4}}f(x)$. Both expressions represent the same mathematical object, so the answer that acknowledges both forms (option D) is correct. The other choices either refer to a different order of derivative or omit the differential operator entirely.

📖 Explanation: When the second derivative is identically zero, the first derivative must be a constant, because differentiating a constant yields zero. Integrating that constant once gives a linear function, and a second integration adds another constant term. Hence the original function must be of the form $f(x)=mx+b$, a linear function.

📖 Explanation: First differentiate: f'(x)=12x^{3}-6x^{2}+2x-4. Differentiating again gives f''(x)=36x^{2}-12x+2. The other options either have incorrect coefficients or miss the factor of three that arises from differentiating a fourth‑degree term twice. Thus option C matches the correct computation.

📖 Explanation: By definition, taking the first derivative yields a first‑order derivative, the second derivative is second‑order, and so on. Each differentiation increments the order by exactly one, so the order of the derivative is identical to the count of differentiations applied. This direct correspondence makes option B the correct description.

📖 Explanation: A higher‑order derivative refers to the derivative obtained after differentiating the original function multiple times, beyond the first derivative. It specifically denotes the nth derivative where $n\ge 2$. Option C captures this idea by stating it is the derivative evaluated at a higher point, which aligns with the standard definition of higher‑order derivatives.

📖 Explanation: Differentiating $e^{kx}$ repeatedly multiplies by the constant factor $k$ each time. After the first derivative we have $k e^{kx}$; after the second derivative $k^{2} e^{kx}$; continuing this pattern yields $g^{(n)}(x)=k^{n}e^{kx}$. The other options either omit the exponentiation of $k$ or introduce extraneous factors, making option A the accurate representation.

📖 Explanation: The second derivative of $f(x)=x^{3}$ is f''(x)=6x, which evaluates to $0$ at $x=0$. For $h(x)=\sin x$, the second derivative is h''(x)=-\sin x, also giving $0$ at $x=0$. Hence both second derivatives vanish at the origin, confirming option D as the correct comparison.

📖 Explanation: An inflection point requires a sign change in the second derivative. For $p(x)$ we find p''(x)=12(x-1)^{2}, which is always non‑negative and equals zero only at $x=1$. However, the sign does not change around $x=1$ because the expression is a perfect square. Therefore the curve has no inflection points, making option C correct.

📖 Explanation: Each differentiation reduces the degree of a polynomial by one. If the fifth derivative vanishes everywhere, the original polynomial must have degree at most four; a degree‑four polynomial becomes zero after five differentiations. Degrees higher than four would produce a non‑zero fifth derivative. Hence the maximum possible degree is four, corresponding to option A.

📖 Explanation: Starting with f'(x)=1/x, differentiate to obtain f''(x)=-1/x^{2}. Differentiating once more yields f'''(x)=2/x^{3}. The sign is positive because the derivative of $-1/x^{2}$ introduces a factor of $-2$ that cancels the existing negative sign. Thus the correct third derivative is $2/x^{3}$, which matches option B.

📖 Explanation: The differential equation f''=-f has characteristic roots $\pm i$, leading to solutions that are linear combinations of $\sin x$ and $\cos x$. Among the options, $\sin x$ directly satisfies the equation because its second derivative is $-\sin x$. Exponential and hyperbolic functions do not fulfill the required relationship, so option A is correct.

📖 Explanation: Applying the product rule repeatedly: h' = e^{x}(x^{2}+2x); h'' = e^{x}(x^{2}+4x+2); h''' = e^{x}(x^{2}+6x+6). Each differentiation adds a term from the derivative of the exponential and from differentiating the polynomial factor. The final expression matches option D.

📖 Explanation: Taylor's theorem states that $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^{n}$. The coefficient of the term $(x-a)^{n}$ is therefore $\frac{f^{(n)}(a)}{n!}$, directly involving the $n$th derivative evaluated at $a$. This makes option A the accurate description.

📖 Explanation: Integrate the constant third derivative: f''(x)=6x+C_{1}. Using f''(0)=0 gives $C_{1}=0$, so f''=6x. Integrate again: f'(x)=3x^{2}+C_{2}; f'(0)=1 yields $C_{2}=1$. Integrate a final time: $f(x)=x^{3}+x+C_{3}$; $f(0)=0$ forces $C_{3}=0$. Evaluating at $x=2$ gives $f(2)=8+2=10$. Hence option B is correct.

📖 Explanation: The function expands as a geometric series: $\frac{1}{1-x}=\sum_{k=0}^{\infty}x^{k}$. The $n$th derivative of $x^{k}$ at zero is zero unless $k=n$, in which case it equals $n!$. Therefore $g^{(n)}(0)=n!$. The other options introduce incorrect signs or factorial offsets, making option A correct.

📖 Explanation: Convexity of a function is guaranteed when its second derivative is non‑negative everywhere. A positive fourth derivative ensures that the second derivative is increasing, but it does not preclude the second derivative from being negative in some intervals. Hence the condition $f^{(4)}(x)>0$ alone does not imply convexity, making the statement false.