Let us proceed in today's post to a couple of applications of the theorem on the differentiation of the inverse function:

$$

\begin{array}{c}

D_{f} = (-\infty, \infty) = f^{-1}(D_{f^{-1}}) \\

\\

f(D_{f}) = (0, \infty) = D_{f^{-1}}

\end{array}

$$

One can easily check that the conditions of the theorem apply and thus for any $x_{0} \in D_{f}$ and for any $y_{0}=f(x_{0}) \in f(D_{f}) \equiv D_{f^{-1}}$ we have:

$$

(f^{-1})'(y) = (lny)' = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{f'(x)} = \frac{1}{(e^{x})'} = \frac{1}{e^{x}} = \frac{1}{y}

$$

where -following the same notation conventions as we did in the development of the theory- $(f^{-1})'(y) = (lny)'$ denotes differentiation with respect to $y$ while $f'(x) = (e^{x})'$ denotes differentiation with respect to $x$. So we finally arrive at: $(lny)' = \frac{1}{y}$ and since the independent variable can be always named at our choice we arrive at the familiar rule:

$$

(lnx)' = \frac{1}{x}

$$

$\bullet$

$$

1 = \frac{dy}{dy} = e^{lny} \frac{d(lny)}{dy} \Rightarrow \frac{d(lny)}{dy} = \frac{1}{e^{lny}} = \frac{1}{y}

$$

and finally, renaming (as usual) the independent variable form $y$ to $x$ we get the familiar relation

$$

(lnx)' \equiv \frac{d(lnx)}{dx} = \frac{1}{x}

$$

$

\begin{array}{l}

D_{f} = [-\frac{\pi}{2}, \frac{\pi}{2}] = f^{-1}(D_{f^{-1}}) \\

\\

f(D_{f}) = [-1, 1] = D_{f^{-1}}

\end{array}

$

Notice that, if displayed in a common set of axis (that is: with the same independent variable $x$) the graphs of $sinx$ and $arcsinx$ are displayed in the following figure:

However, we will proceed the computation using the notation $y= f(x) = sinx$ and $x = f^{-1}(y) = arcsiny$.

We first have to note that the theorem provides us the derivative of the inverse function $x = f^{-1}(y) = arcsiny$ for any $y \in (-1,1)$ excluding thus the points $sin(-\frac{\pi}{2})=-1$, $sin(\frac{\pi}{2})=1$ simply because $f'(x) = (sinx)' = cosx$ and thus $f'(-\frac{\pi}{2})=f'(\frac{\pi}{2})=0$. However, the conditions of the theorem are valid in $(-1,1)$ so we get:

$$

\begin{array}{c}

(f^{-1})'(y) = (arcsiny)' = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{f'(x)} = \frac{1}{(sinx)'} = \\

\\

= \frac{1}{cosx} = \frac{1}{\sqrt{1-sin^{2}x}} = \frac{1}{\sqrt{1-y^{2}}}

\end{array}

$$

for $y$ in $(-1,1)$. Notice that we have used $cosx = \sqrt{1-sin^{2}x} \geq 0$ for $x \in (-\frac{\pi}{2},\frac{\pi}{2})$. We thus have shown (after renaming the independent variable as is customary) that:

$$

(arcsinx)' = \frac{1}{\sqrt{1-x^{2}}}

$$

for any $x$ in $(-1,1)$.

$\bullet$

$$

\begin{array}{c}

1 = \frac{dy}{dy} = cos(arcsiny) \frac{d(arcsiny)}{dy} \Rightarrow \\

\\

\Rightarrow \frac{d(arcsiny)}{dy} = \frac{1}{cos(arcsiny)} = \frac{1}{\sqrt{1-sin^{2}(arcsiny)}} = \frac{1}{\sqrt{1-y^{2}}}

\end{array}

$$

for $y$ in $(-1,1)$. With the -customary now- change in the independent variable from $y$ to $x$ we arrive at our desired formula:

$$

(arcsinx)' = \frac{1}{\sqrt{1-x^{2}}}

$$

for any $x$ in $(-1,1)$.

**$\bullet$ Let us consider the case of $y=f(x)=e^{x}$. It is well known that $f'(x)=e^{x}=f(x)$ and that the inverse function $f^{-1}$ can be written as: $x=f^{-1}(y)=lny$. The domain and the range of these functions are**__Example 1 (exp-log):__$$

\begin{array}{c}

D_{f} = (-\infty, \infty) = f^{-1}(D_{f^{-1}}) \\

\\

f(D_{f}) = (0, \infty) = D_{f^{-1}}

\end{array}

$$

One can easily check that the conditions of the theorem apply and thus for any $x_{0} \in D_{f}$ and for any $y_{0}=f(x_{0}) \in f(D_{f}) \equiv D_{f^{-1}}$ we have:

$$

(f^{-1})'(y) = (lny)' = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{f'(x)} = \frac{1}{(e^{x})'} = \frac{1}{e^{x}} = \frac{1}{y}

$$

where -following the same notation conventions as we did in the development of the theory- $(f^{-1})'(y) = (lny)'$ denotes differentiation with respect to $y$ while $f'(x) = (e^{x})'$ denotes differentiation with respect to $x$. So we finally arrive at: $(lny)' = \frac{1}{y}$ and since the independent variable can be always named at our choice we arrive at the familiar rule:

$$

(lnx)' = \frac{1}{x}

$$

$\bullet$

__Let us see how we could have work alternatively__, thinking through the chain rule of differentiating composite function: We could differentiate the composition $f\circ f^{-1}$ with respect to (the independent variable) $y$. So we get $y = e^{x} = e^{x(y)} = e^{lny}$. Differentiating both sides with respect to $y$ and applying the chain rule in the r.h.s of the last relation, we get$$

1 = \frac{dy}{dy} = e^{lny} \frac{d(lny)}{dy} \Rightarrow \frac{d(lny)}{dy} = \frac{1}{e^{lny}} = \frac{1}{y}

$$

and finally, renaming (as usual) the independent variable form $y$ to $x$ we get the familiar relation

$$

(lnx)' \equiv \frac{d(lnx)}{dx} = \frac{1}{x}

$$

__Example 2 (inverse trigonometric functions):__$\bullet$ Let $y=f(x)=sinx$. Its inverse function can be written as $x=f^{-1}(y)=arcsin(y) \equiv sin^{-1}(x)$. Of course $sinx$ when considered in its natural domain $\mathbb{R}$ is not a $``1-1"$ and thus not an invertible function. However a suitable restriction is: We consider the domains and ranges as follows:$

\begin{array}{l}

D_{f} = [-\frac{\pi}{2}, \frac{\pi}{2}] = f^{-1}(D_{f^{-1}}) \\

\\

f(D_{f}) = [-1, 1] = D_{f^{-1}}

\end{array}

$

Notice that, if displayed in a common set of axis (that is: with the same independent variable $x$) the graphs of $sinx$ and $arcsinx$ are displayed in the following figure:

However, we will proceed the computation using the notation $y= f(x) = sinx$ and $x = f^{-1}(y) = arcsiny$.

We first have to note that the theorem provides us the derivative of the inverse function $x = f^{-1}(y) = arcsiny$ for any $y \in (-1,1)$ excluding thus the points $sin(-\frac{\pi}{2})=-1$, $sin(\frac{\pi}{2})=1$ simply because $f'(x) = (sinx)' = cosx$ and thus $f'(-\frac{\pi}{2})=f'(\frac{\pi}{2})=0$. However, the conditions of the theorem are valid in $(-1,1)$ so we get:

$$

\begin{array}{c}

(f^{-1})'(y) = (arcsiny)' = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{f'(x)} = \frac{1}{(sinx)'} = \\

\\

= \frac{1}{cosx} = \frac{1}{\sqrt{1-sin^{2}x}} = \frac{1}{\sqrt{1-y^{2}}}

\end{array}

$$

for $y$ in $(-1,1)$. Notice that we have used $cosx = \sqrt{1-sin^{2}x} \geq 0$ for $x \in (-\frac{\pi}{2},\frac{\pi}{2})$. We thus have shown (after renaming the independent variable as is customary) that:

$$

(arcsinx)' = \frac{1}{\sqrt{1-x^{2}}}

$$

for any $x$ in $(-1,1)$.

$\bullet$

__Let us now try to work alternatively__(through the chain rule) as before: We differentiate $f\circ f^{-1}$ with respect to (the independent variable) $y$ using the chain rule and keeping in mind that $y = sinx \equiv sinx(y) = sin(arcsin y)$$$

\begin{array}{c}

1 = \frac{dy}{dy} = cos(arcsiny) \frac{d(arcsiny)}{dy} \Rightarrow \\

\\

\Rightarrow \frac{d(arcsiny)}{dy} = \frac{1}{cos(arcsiny)} = \frac{1}{\sqrt{1-sin^{2}(arcsiny)}} = \frac{1}{\sqrt{1-y^{2}}}

\end{array}

$$

for $y$ in $(-1,1)$. With the -customary now- change in the independent variable from $y$ to $x$ we arrive at our desired formula:

$$

(arcsinx)' = \frac{1}{\sqrt{1-x^{2}}}

$$

for any $x$ in $(-1,1)$.

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