Question

Find the derivatives of each of the functions in Exercises 17–50. In some cases it may be convenient to do some preliminary algebra

$f\left(x\right)=x^{-2}{e}^{4x}{\sin }^{-1}x$

Short answer

The derivative is$\frac{4e^{4x}{\sin }^{-1}x}{x^2}+\frac{e^{4x}}{x^2\sqrt{1-x^2}}-\frac{2e^{4x}{\sin }^{-1}x}{x^3}$.

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)=x^{-2}{e}^{4x}{\sin }^{-1}x$.

Step 2. Preliminary Algebra.

We know,

Quotient Rule:

${\left(\frac{f}{g}\right)}^{\text{'}}\left(x\right)=\frac{f^{\text{'}}\left(x\right)g\left(x\right)-f\left(x\right)g^{\text{'}}\left(x\right)}{\left(g\right(x){)}^2}$

Product rule:

$\left(fg{)}^{\text{'}}\right(x)=f^{\text{'}}(x\left)g\right(x)+f(x\left)g^{\text{'}}\right(x)$

Chain Rule:

$(f\circ u{)}^{\text{'}}(x)=f^{\text{'}}(u\left(x\right)\left)u^{\text{'}}\right(x)$

$\frac{d}{dx}\left({\sin }^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}{e}^x=e^x$

Step 3. Derivative of the function.

The derivative of the function will be,

$$\begin{gathered} \frac{d}{dx}\left(x^{-2}{e}^{4x}{\sin }^{-1}x\right)=\frac{d}{dx}\left(\frac{e^{4x}{\sin }^{-1}x}{x^2}\right) \\ =\left(\frac{\left(\frac{d}{dx}\left(e^{4x}{\sin }^{-1}x\right)\right)x^2-\left(e^{4x}{\sin }^{-1}x\right)\frac{d}{dx}\left(x^2\right)}{{\left(x^2\right)}^2}\right) \\ =\left(\frac{\left(\left(\frac{d}{dx}{e}^{4x}\right)\left({\sin }^{-1}x\right)+\left(\frac{d}{dx}{\sin }^{-1}x\right)\left(e^{4x}\right)\right)x^2-\left(2x{e}^{4x}{\sin }^{-1}x\right)}{x^4}\right) \\ =\left(\frac{\left(e^{4x}\left(\frac{d}{dx}4x\right)\left({\sin }^{-1}x\right)+\left(\frac{1}{\sqrt{1-x^2}}\right)\left(e^{4x}\right)\right)x^2-\left(2x{e}^{4x}{\sin }^{-1}x\right)}{x^4}\right) \\ =\left(\frac{\left(4e^{4x}{\sin }^{-1}x+\left(\frac{1}{\sqrt{1-x^2}}\right)\left(e^{4x}\right)\right)x^2-\left(2x{e}^{4x}{\sin }^{-1}x\right)}{x^4}\right) \\ =\left(\frac{x^2\left(4e^{4x}{\sin }^{-1}x+\frac{e^{4x}}{\sqrt{1-x^2}}\right)-\left(2x{e}^{4x}{\sin }^{-1}x\right)}{x^4}\right) \\ =\left(\frac{x^2\left(4e^{4x}{\sin }^{-1}x+\frac{e^{4x}}{\sqrt{1-x^2}}\right)}{x^4}-\left(\frac{\left(2x{e}^{4x}{\sin }^{-1}x\right)}{x^4}\right)\right) \\ =\frac{4e^{4x}{\sin }^{-1}x}{x^2}+\frac{e^{4x}}{x^2\sqrt{1-x^2}}-\frac{2e^{4x}{\sin }^{-1}x}{x^3} \end{gathered}$$