式展開: 2を底とする指数凾数の微分

\begin{array}{rcl} y & = & 2^{x} \end{array}

\begin{array}{rcl} \ln (y) & = & \ln (2^{x}) \\ e^{\ln (y)} & = & e^{\ln (2^{x})} \\ y & = & e^{x \ln (2)} \dots \ln (A^{B}) = B \ln (A) \end{array}

\begin{array}{rcl} x & = & \log_{2} (y) \\ = & \frac{\ln (y)}{\ln (2)} \dots \log_{A} (B) = \frac{\log_{C} (B)}{\log_{C} (A)} \\ \ln (y) & = & x \ln (2) \\ e^{\ln (y)} & = & e^{x \ln (2)} \\ y & = & e^{x \ln (2)} \end{array}

\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d}{d x} 2^{x} \\ = & \frac{d}{d x} e^{x \ln (2)} \\ = & \frac{d e^{u}}{d u} \frac{d u}{d x} \dots u = x \ln (2) \\ = & e^{u} \cdot \ln (2) \dots \frac{d e^{x}}{d x} = e^{x}, \frac{d u}{d x} = \frac{d}{d x} x \ln (2) = \ln (2) \\ = & e^{x \ln (2)} \ln (2) \\ = & e^{\ln (2^{x})} \ln (2) \\ = & 2^{x} \ln (2) \end{array}

\begin{array}{rcl} y & = & a^{x} \end{array}

\begin{array}{rcl} \ln (y) & = & \ln (a^{x}) \\ e^{\ln (y)} & = & e^{\ln (a^{x})} \\ y & = & e^{x \ln (a)} \dots \ln (A^{B}) = B \ln (A) \end{array}

\begin{array}{rcl} x & = & \log_{a} (y) \\ = & \frac{\ln (y)}{\ln (a)} \dots \log_{A} (B) = \frac{\log_{C} (B)}{\log_{C} (A)} \\ \ln (y) & = & x \ln (a) \\ e^{\ln (y)} & = & e^{x \ln (a)} \\ y & = & e^{x \ln (a)} \end{array}

\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d}{d x} a^{x} \\ = & \frac{d}{d x} e^{x \ln (a)} \\ = & \frac{d e^{u}}{d u} \frac{d u}{d x} \dots u = x \ln (a) \\ = & e^{u} \cdot \ln (a) \dots \frac{d e^{x}}{d x} = e^{x}, \frac{d u}{d x} = \frac{d}{d x} x \ln (a) = \ln (a) \\ = & e^{x \ln (a)} \ln (a) \\ = & e^{\ln (a^{x})} \ln (a) \\ = & a^{x} \ln (a) \end{array}

式展開