Initialize

Loading Student:-Calculus1

Solution #1

${\mathrm{sech}}^{-1}\left(1\/x\right)$$\stackrel{\text{differentiate w.r.t. x}}{\to }$$\frac{1}{\sqrt{x-1}\sqrt{x\+1}}$$$

Solution #2

$\frac{\ⅆ}{\ⅆ x} {\mathrm{sech}}^{-1}\left(1\/x\right)$ = $\frac{1}{\sqrt{x-1}\sqrt{x\+1}}$$$

Solution #3 (stepwise)

$$$\frac{\ⅆ}{\ⅆ x} {\mathrm{sech}}^{-1}\left(1\/x\right)$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Differentiation Steps}\\ & & \frac{\ⅆ}{\ⅆx}\mathrm{arcsech}\left(\frac{1}{x}\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{chain}}\text{rule to the term}\mathrm{arcsech}\left(\frac{1}{x}\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{chain}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)=\mathrm{f\text{'}}\left(g\left(x\right)\right)\frac{\ⅆ}{\ⅆx}g\left(x\right)\\ & \text{◦}& \text{Outside function}\\ & & f\left(v\right)=\mathrm{arcsech}\left(v\right)\\ & \text{◦}& \text{Inside function}\\ & & g\left(x\right)=\frac{1}{x}\\ & \text{◦}& \text{Derivative of outside function}\\ & & \frac{\ⅆ}{\ⅆv}f\left(v\right)=-\frac{1}{v^2\sqrt{\frac{1}{v}-1}\sqrt{\frac{1}{v}+1}}\\ & \text{◦}& \text{Apply composition}\\ & & \mathrm{f\text{'}}\left(g\left(x\right)\right)=-\frac{x^2}{\sqrt{x-1}\sqrt{x+1}}\\ & \text{◦}& \text{Derivative of inside function}\\ & & \frac{\ⅆ}{\ⅆx}g\left(x\right)=-\frac{1}{x^2}\\ & \text{◦}& \text{Put it all together}\\ & & \frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)\frac{\ⅆ}{\ⅆx}g\left(x\right)=\left[\right]\\ & & \text{This gives:}\\ & & \frac{1}{\sqrt{x-1}\sqrt{x+1}}\end{array}$$$

Figure 2.10.4(a)   Context Panel access to the Differentiation Rules