📝 Constant multiple rule derivative (14 MCQs)

📖 Explanation: The rule that permits moving a constant through the derivative sign is precisely the Constant Multiple Rule. It states that for any constant $c$ and differentiable function $f$, the derivative of $c\,f(x)$ equals c\,f'(x). This principle is the foundation for all subsequent examples.

📖 Explanation: For $f(x)=x^{2}$ the derivative is $2x$. Multiplying the original function by 5 yields $5x^{2}$; applying the Constant Multiple Rule gives derivative $5\cdot2x =10x$. Hence the derivative of $5x^{2}$ is exactly five times the derivative of $x^{2}$.

📖 Explanation: Given $g(x) = -2h(x)$ and h'(2)=4, the Constant Multiple Rule tells us to multiply the known derivative by the constant factor: g'(2)= -2 \times 4 = -8. The sign and magnitude are directly inherited from the constant multiplier.

📖 Explanation: A constant function $c$ does not change with $x$; its rate of change is zero. Therefore $\frac{d}{dx}[c]=0$ regardless of the value of $c$. This follows directly from the definition of derivative as a limit of a zero difference.

📖 Explanation: Since $\sin x$ differentiates to $\cos x$, the function $7\sin x$ has derivative $7\cos x$ by the Constant Multiple Rule. The constant 7 remains outside the differentiation, and only the inner function $\sin x$ contributes its derivative.

📖 Explanation: The derivative of $4p(x)$ equals 4p'(x). If at $x=2$ this derivative equals 96, then 4p'(2)=96 implying p'(2)=96/4=24. This logical step isolates the unknown p'(2) using the constant factor.

📖 Explanation: The Constant Multiple Rule applies only when the constant multiplies the entire function. For $e^{2x}$ the exponent contains the variable, so differentiating requires the Chain Rule, not just the constant factor rule. Hence this function cannot be handled solely by the constant multiple rule.

📖 Explanation: Using the rule, the derivative of $c\ln x$ is $c\cdot\frac{1}{x}= \frac{c}{x}$. For $\ln(x^{c})$, rewrite as $c\ln x$ before differentiating, which yields the same result $\frac{c}{x}$. Thus both derivatives are identical.

📖 Explanation: From $q(x)=k f(x)$ we have q'(a)=k f'(a). Substituting the given numbers gives $12 = k\cdot3$, so solving for $k$ yields $k=4$. The constant factor is directly determined by dividing the known derivative values.

📖 Explanation: The rule holds for any real constant, positive, negative, or irrational, because the limit definition treats the constant as a scalar that can be factored out of the difference quotient. Whether the constant is $-\pi$ or $\sqrt{2}$ does not affect the algebraic step of moving it through the limit.

📖 Explanation: First rewrite $\frac{1}{5}x^{5}$ as $\frac{1}{5}\cdot x^{5}$. Applying the Constant Multiple Rule gives derivative $\frac{1}{5}\cdot5x^{4}=x^{4}$. The intermediate option $\frac{1}{5}5x^{4}$ simplifies to the same result, confirming the correct choice.

📖 Explanation: Given f'(x)=2f(x), the derivative of $3f(x)$ is 3f'(x)=3\cdot2f(x)=6f(x) by the Constant Multiple Rule. This shows that the derivative of the scaled function is six times the original function value, preserving the proportional relationship.

📖 Explanation: For $u(x)=\sqrt{2}\,(x^{3}-4x)$, the first derivative is $\sqrt{2}(3x^{2}-4)$. Differentiating again, the constant $\sqrt{2}$ stays outside, and the derivative of $3x^{2}-4$ is $6x$. Hence u''(x)=\sqrt{2}\cdot6x=6\sqrt{2}x.

📖 Explanation: The student's claim ignores the derivative of the inner function; the constant multiple rule multiplies the constant by the derivative of $f(x)$, not by the function itself. For $2\sin x$, the correct derivative is $2\cos x$. The error lies in assuming the derivative equals the constant alone.