3.9 Derivatives of exponential and logarithmic functions  (Page 3/6)

Proof

If $x>0$ and $y=\text{ln}x,$ then $e^y=x.$ Differentiating both sides of this equation results in the equation

Solving for $\frac{dy}{dx}$ yields

Finally, we substitute $x=e^y$ to obtain

We may also derive this result by applying the inverse function theorem, as follows. Since $y=g(x)=\text{ln}x$ is the inverse of $f\left(x\right)=e^x,$ by applying the inverse function theorem we have

Using this result and applying the chain rule to $h\left(x\right)=\text{ln}\left(g\left(x\right)\right)$ yields

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The graph of $y=\text{ln}x$ and its derivative $\frac{dy}{dx}=\frac{1}{x}$ are shown in [link] .

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of $y=lo{g}_{b}x$ and $y=b^x$ for $b>0,b\ne 1.$

Proof

If $y={\text{log}}_{b}x,$ then $b^y=x.$ It follows that $\text{ln}\left(b^y\right)=\text{ln}x.$ Thus $y\text{ln}b=\text{ln}x.$ Solving for $y,$ we have $y=\frac{\text{ln}x}{\text{ln}b}.$ Differentiating and keeping in mind that $\text{ln}b$ is a constant, we see that

The derivative in [link] now follows from the chain rule.

If $y=b^x,$ then $\text{ln}y=x\text{ln}b.$ Using implicit differentiation, again keeping in mind that $\text{ln}b$ is constant, it follows that $\frac{1}{y}\frac{dy}{dx}=\text{ln}b.$ Solving for $\frac{dy}{dx}$ and substituting $y=b^x,$ we see that

The more general derivative ( [link] ) follows from the chain rule.

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Logarithmic differentiation

At this point, we can take derivatives of functions of the form $y={\left(g\left(x\right)\right)}^n$ for certain values of $n,$ as well as functions of the form $y=b^{g(x)},$ where $b>0$ and $b\ne 1.$ Unfortunately, we still do not know the derivatives of functions such as $y=x^x$ or $y=x^{\pi }.$ These functions require a technique called logarithmic differentiation    , which allows us to differentiate any function of the form $h\left(x\right)=g{\left(x\right)}^{f(x)}.$ It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of $y=\frac{x\sqrt{2x+1}}{e^x{\text{sin}}^{3}x}.$ We outline this technique in the following problem-solving strategy.