Chapter 3.6, Problem 30E

To determine

To calculate: The derivative of the function $f\left(x\right)=\arcsin \left(e^x\right)$ . Also evaluate the domain of the function and its derivative.

Answer to Problem 30E

The derivative of the function is $f\text{'}\left(x\right)=\frac{e^x}{\sqrt{1-e^{2x}}}$ . The domain of the function is $x\in \left(-\infty ,0\right]$ and domain of derivative of the function is $x\in \left(-\infty ,0\right)$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=\arcsin \left(e^x\right)$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Derivative of inverse trigonometric function $\frac{d}{dx}\left(\arcsin \left(x\right)\right)=\frac{1}{\sqrt{1-x^2}}$ .

Calculation:

Consider the function $f\left(x\right)=\arcsin \left(e^x\right)$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(\arcsin \left(e^x\right)\right)\\ f\text{'}\left(x\right)=\frac{d}{dx}\left(\arcsin \left(e^x\right)\right)\end{array}$

Recall that thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Derivative of inverse trigonometric function $\frac{d}{dx}\left(\arcsin \left(x\right)\right)=\frac{1}{\sqrt{1-x^2}}$ .

Apply it,

  $\begin{array}{c}f\text{'}\left(x\right)=\frac{1}{\sqrt{1-{\left(e^x\right)}^2}}\frac{d}{dx}\left(e^x\right)\\ f\text{'}\left(x\right)=\frac{e^x}{\sqrt{1-{\left(e^x\right)}^2}}\\ f\text{'}\left(x\right)=\frac{e^x}{\sqrt{1-e^{2x}}}\end{array}$

Thus, the derivative of the function is $f\text{'}\left(x\right)=\frac{e^x}{\sqrt{1-e^{2x}}}$ .

Consider the function, $f\text{'}\left(x\right)=\frac{e^x}{\sqrt{1-e^{2x}}}$ .

Recall that domain of the function is the set of values of independent variable for which function is defined.

The function of the form $\frac{p}{q}$ fraction are defined when $q\ne 0$ that is denominator should not be equal to 0.Also, the square root sign is defined for positive values.

The provided expression $f\text{'}\left(x\right)=\frac{e^x}{\sqrt{1-e^{2x}}}$ is defined for real values that is, $1-e^{2x}>0$ , therefore,

  $\begin{array}{c}{e}^{2x}<1\\ \ln e^{2x}<\ln 1\\ 2x<0\\ x<0\end{array}$

Therefore, domain is $x\in \left(-\infty ,0\right)$ for function to be defined.

Consider the function, $f\left(x\right)=\arcsin \left(e^x\right)$ .

Recall that domain of the function is the set of values of independent variable for which function is defined.

The domain of inverse sine trigonometric function is $\left[-1,1\right]$ . Also the domain of exponential function is all real numbers. For the domain of $e^x$ to lie in $\left[-1,1\right]$ the value of x should be less than or equal to 0.

Therefore, domain is $x\in \left(-\infty ,0\right]$ for function to be defined.

Hence, the derivative of the function is $f\text{'}\left(x\right)=\frac{e^x}{\sqrt{1-e^{2x}}}$ . The domain of the function is $x\in \left(-\infty ,0\right]$ and domain of derivative of the function is $x\in \left(-\infty ,0\right)$ .