Module 1: Now You Try I

Differentiate the following functions:

1. $f (x) = \ln (\sqrt[3]{x^{2} + 1} - e^{x^{2}})$ $f(x)\;=\; \ln(\sqrt[3]{x^2+1}-e^{x^2})$

Solution:

▶ Show

f^{'} (x) = \frac{\frac{1}{3} (x^{2} + 1)^{\frac{- 2}{3}} (2 x) - (2 x) e^{x^{2}}}{\sqrt[3]{x^{2} + 1} - e^{x^{2}}}

$f'(x)=\frac{\frac{1}{3}({x^2+1)^{\frac{-2}{3}}(2x)-(2x)e^{x^2}}}{\sqrt[3]{x^2+1}-e^{x^2}}$

2. $g (x) = \sin^{(n - 1)} (\frac{1}{x})$ $g(x)\;=\; \sin^{(n-1)}\left(\frac{1}{x}\right)$

Solution:

▶ Show

g^{'} (x) = \frac{- 1}{x^{2}} (n - 1) \sin^{(n - 2)} (\frac{1}{x}) \cos (\frac{1}{x})

$g'(x)=\frac{-1}{x^2}(n-1)\sin^{(n-2)}\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right)$

3. If $y$ $y$ is a function of $x$ $x$ , compute ${(\frac{x}{x - y^{3}})}^{'}$ $\left(\frac{x}{x-y^3}\right)'$ and $LaTeX: \left(\sin^3\left(\frac{x}{x-y^3}\right)\right)'$ $\left(\sin^3\left(\frac{x}{x-y^3}\right)\right)'$ .

Solution:

▶ Show

{(\frac{x}{x - y^{3}})}^{'} = \frac{1 (x - y^{3}) - x (1 - 3 y^{2} y^{'})}{(x - y^{3})^{2}}

$\displaystyle \left(\frac{x}{x-y^3}\right)'=\frac{1(x-y^3)-x(1-3y^2y')}{(x-y^3)^2}$ and

$LaTeX: \displaystyle \left(\sin^3\left(\frac{x}{x-y^3}\right)\right)'=3\sin^2\left(\frac{x}{x-y^3}\right)\cos\left(\frac{x}{x-y^3}\right)\frac{1(x-y^3)-x(1-3y^2y')}{(x-y^3)^2}$

$\displaystyle \left(\sin^3\left(\frac{x}{x-y^3}\right)\right)'=3\sin^2\left(\frac{x}{x-y^3}\right)\cos\left(\frac{x}{x-y^3}\right)\frac{1(x-y^3)-x(1-3y^2y')}{(x-y^3)^2}$