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If $x>0$ and $y=\text{ln}\phantom{\rule{0.1em}{0ex}}x,$ 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 $y=g(x)=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ 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 $h\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right)$ yields
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The graph of $y=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ and its derivative $\frac{dy}{dx}=\frac{1}{x}$ are shown in [link] .
Find the derivative of $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}+3x-4\right).$
Use [link] directly.
Find the derivative of $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\frac{{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x}{2x+1}\right).$
At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.
Differentiate: $f\left(x\right)=\text{ln}{\left(3x+2\right)}^{5}.$
${f}^{\prime}\left(x\right)=\frac{15}{3x+2}$
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 $b>0,b\ne 1.$
Let $b>0,b\ne 1,$ and let $g\left(x\right)$ be a differentiable function.
If $y={\text{log}}_{b}x,$ then ${b}^{y}=x.$ It follows that $\text{ln}\phantom{\rule{0.1em}{0ex}}\left({b}^{y}\right)=\text{ln}\phantom{\rule{0.2em}{0ex}}x.$ Thus $y\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}b=\text{ln}\phantom{\rule{0.2em}{0ex}}x.$ Solving for $y,$ we have $y=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{ln}\phantom{\rule{0.1em}{0ex}}b}.$ Differentiating and keeping in mind that $\text{ln}\phantom{\rule{0.1em}{0ex}}b$ is a constant, we see that
The derivative in [link] now follows from the chain rule.
If $y={b}^{x},$ then $\text{ln}\phantom{\rule{0.2em}{0ex}}y=x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b.$ Using implicit differentiation, again keeping in mind that $\text{ln}\phantom{\rule{0.1em}{0ex}}b$ is constant, it follows that $\frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}=\text{ln}\phantom{\rule{0.1em}{0ex}}b.$ 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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Find the derivative of $h\left(x\right)=\frac{{3}^{x}}{{3}^{x}+2}.$
Use the quotient rule and [link] .
Find the slope of the line tangent to the graph of $y={\text{log}}_{2}\left(3x+1\right)$ at $x=1.$
To find the slope, we must evaluate $\frac{dy}{dx}$ at $x=1.$ Using [link] , we see that
By evaluating the derivative at $x=1,$ we see that the tangent line has slope
Find the slope for the line tangent to $y={3}^{x}$ at $x=2.$
$9\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(3\right)$
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.
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