3.9 Derivatives of exponential and logarithmic functions  (Page 2/6)

See the following table.

The evidence from the table suggests that $2.7182<e<2.7183.$

The graph of $E\left(x\right)=e^x$ together with the line $y=x+1$ are shown in [link] . This line is tangent to the graph of $E\left(x\right)=e^x$ at $x=0.$

Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of $B\left(x\right)=b^x,b>0.$ Recall that we have assumed that $B^{\prime }\left(0\right)$ exists. By applying the limit definition to the derivative we conclude that

Turning to $B^{\prime }\left(x\right),$ we obtain the following.

We see that on the basis of the assumption that $B\left(x\right)=b^x$ is differentiable at $0,B\left(x\right)$ is not only differentiable everywhere, but its derivative is

For $E\left(x\right)=e^x,E^{\prime }\left(0\right)=1.$ Thus, we have $E^{\prime }\left(x\right)=e^x.$ (The value of $B^{\prime }\left(0\right)$ for an arbitrary function of the form $B\left(x\right)=b^x,b>0,$ will be derived later.)

Derivative of the logarithmic function

Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.