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NSW Y12 Maths - Advanced Exponential and Log Function Applications of Differentiation e^x

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Applications of Differentiation e^x Theory

Applications of the differentiation of exponential functions involves determining equations of tangents and stationary points on these functions.\\ \begin{multicols}{2} \textbf{Example 1}\\ %10634 Find the equation of the tangent to the curve \(y = {e^x}{(x - 1)^2}\) at the point where \(x = 2\) \\ \textbf{Example 1 solution}\\ $\begin{aligned} y &=e^{x}(x-1)^{2} \\ \text { let } u &=e^{x} \quad v=(x-1)^{2} \\ d u &=e^{x} \quad d v=2(x-1) \\ y^{\prime} &=u d v+v d e \\ &=2 e^{x}(x-1)+e^{x}(x-1)^{2}\\ \text{At}\ x&=2\\ y^{\prime} &=2 e^{2}+e^{2} \\ &=3 e^{2} \\ y &=e^{2} \\ y-e^{2}&=3 e^{2}(x-2)\\ y-e^{2}&=3 e^{2} x-6 e^{2}\\ 3 e^{2} x-y-5 e^{2}&=0 \end{aligned}$\\ \columnbreak \textbf{Example 2}\\%10635 Find the stationary point on the curve \(y = x{e^{ - x}}\) and determine its nature.\\ \textbf{Example 2 solution}\\ $\begin{aligned} y &=x e^{-x} \\ \text { let } u &=x \quad v=e^{-x} \\ d u &=1 \quad d v=-e^{-x} \\ y^{\prime} &=u d v+v d u \\ &=-x e^{-x}+e^{-x}\\ y^{\prime \prime} &=-\left(-x e^{-x}+e^{-x}\right)-e^{-x} \\ &=x e^{-x}-2 e^{-x}\\ \text { let } y^{\prime}&=0 \quad e^{-x}(-x+1)=0\\ \therefore\ x&=1, y=e^{-1}\\ \text{At}\ x=1 \quad y^{\prime \prime}&=e^{-1}-2 e^{-1}=-\frac{1}{e} \rightarrow y^{\prime \prime}<0\\ \therefore\ \text{max at}\ &\left(1, \frac{1}{e}\right) \end{aligned}$ \end{multicols}

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