Cards In This Deck

• $$\begin{gathered} \text{True or False?} \\ \text{The derivative of }{3}^x\text{ is }x{3}^{x-1}\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }{3}^x\ln 3 \\ {x}^n\text{ is a power function (variable base, constant power). }{a}^x\text{ is an exponential function (constant base, variable power). They follow different rules.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \frac{d}{dx}(3e^{2x})=3e^{2x} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }6e^{2x} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \frac{d}{dx}(e^{5x})=e^{5x} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }5e^{5x} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{To find the gradient of the curve }y=\sin (2x)\text{ at }x=30\text{, you calculate }2\cos (60)\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Method: You must convert }x\text{ to radians before calculating.} \\ \text{Calculus with trigonometric functions only works when angles are measured in radians. }x=30\text{ would be treated as 30 radians unless degrees are explicitly handled (which is non-standard in calculus questions).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{The derivative of }\cot (x)\text{ is }{cosec}^2(x)\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }-{cosec}^2(x) \\ \text{All ’co-’ trigonometric functions (cos, cot, cosec) have a negative sign in their derivatives.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \frac{d}{dx}(\ln (5x))=\frac{5}{5x}=\frac{1}{x} \end{gathered}$$
  $$\begin{gathered} \text{✅ True} \\ \text{However, a common error is to think the answer is }\frac{5}{x}\text{. } \\ \text{Using the chain rule: }\frac{d}{dx}(\ln (f(x)))=\frac{f^{\prime }(x)}{f(x)}\text{. Here, }\frac{5}{5x}\text{ simplifies to }\frac{1}{x}\text{. Alternatively, }\ln (5x)=\ln 5+\ln x\text{, and since }\ln 5\text{ is a constant, the derivative is just }\frac{1}{x}\text{.} \end{gathered}$$
• $\text{For the function }y=\frac{1}{\cos x}\text{, how can you derive the derivative of }\sec x\text{ without memorizing the formula?}$
  $$\begin{gathered} \text{Two methods:} \\ \text{1. Chain Rule: Write as }y=(\cos x{)}^{-1}\text{. Differentiate to get }-1(\cos x{)}^{-2}\times (-\sin x)\text{.} \\ \text{2. Quotient Rule: Set }u=1\text{, }v=\cos x\text{.} \end{gathered}$$
• $\text{When differentiating }y=\frac{e^{3x}}{x^2}\text{, is it better to use the Quotient Rule or Product Rule?}$
  $$\begin{gathered} \text{Both work, but:} \\ \text{1. Quotient Rule is standard for fractions.} \\ \text{2. Product Rule works if you rewrite as }y=e^{3x}{x}^{-2}\text{.} \\ \text{Decision: If the denominator is a single term (like }{x}^2\text{), rewriting and using the Product Rule is often less prone to algebraic simplification errors.} \end{gathered}$$
• $\text{Which differentiation rule should you primarily use for }y={\sin }^3(x)\text{ versus }y=x^3\sin (x)\text{?}$
  $$\begin{gathered} \text{1. }y={\sin }^3(x)\text{: Use the Chain Rule (Function of a function: }(\sin x{)}^3\text{).} \\ \text{2. }y=x^3\sin (x)\text{: Use the Product Rule (Two separate functions of }x\text{ multiplied: }u=x^3\text{, }v=\sin x\text{).} \end{gathered}$$
• $\text{What is the procedure to find the x-coordinate of a stationary point for the curve }y=x{e}^{-2x}\text{?}$
  $$\begin{gathered} \text{1. Use the Product Rule to find }\frac{dy}{dx}\text{.} \\ \text{2. Set }\frac{dy}{dx}=0\text{.} \\ \text{3. Factorise out the exponential term (e.g., }{e}^{-2x}\text{).} \\ \text{4. Since }{e}^{-2x}\ne 0\text{, solve the remaining linear/algebraic part.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{When using the Quotient Rule for }y=\frac{u}{v}\text{, the formula is }\frac{dy}{dx}=\frac{u^{\prime }v-v^{\prime }u}{v^2}\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Formula: }\frac{dy}{dx}=\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2} \\ \text{The order matters because of the subtraction in the numerator. You must start with }v\text{ times the derivative of }u\text{.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \frac{d}{dx}({\cos }^2(x))=-{\sin }^2(x) \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }-2\cos (x)\sin (x)\text{ (or }-\sin (2x)\text{)} \\ \text{This is a Chain Rule problem. Let }u=\cos x\text{, then }y=u^2\text{. } \\ \frac{dy}{dx}=2u\times \frac{du}{dx}=2(\cos x)\times (-\sin x)\text{.} \end{gathered}$$
• $\text{What is the first step to find }\frac{dy}{dx}\text{ if you are given an equation in the form }x=\sin (2y)\text{?}$
  $$\begin{gathered} \text{Differentiate with respect to }y\text{ to find }\frac{dx}{dy}\text{ first, then take the reciprocal.} \\ \frac{dx}{dy}=2\cos (2y)⟹\frac{dy}{dx}=\frac{1}{2\cos (2y)} \end{gathered}$$
• $\text{When proving the derivative of }y=\arcsin x\text{, what is the standard starting step?}$
  $$\begin{gathered} \text{Rearrange the equation to remove the inverse function:} \\ x=\sin y \\ \text{Then differentiate implicitly with respect to }x\text{ (or find }dx/dy\text{ and flip).} \end{gathered}$$
• $$\begin{gathered} \text{A population }P\text{ is modelled by }P=P_0{e}^{kt}\text{. } \\ \text{How do you find the initial population from the equation?} \end{gathered}$$
  $\text{Substitute }t=0\text{ into the equation.}$
• $$\begin{gathered} \text{In a modelling question, you derive the differential equation }\frac{dP}{dt}=kP\text{.} \\ \text{What is the general form of the solution for }P\text{ in terms of }t\text{?} \end{gathered}$$
  $P=A{e}^{kt}$