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Differentiate the following. Simplify, as appropriate.
$\displaystyle{y = \frac{x}{\cot x}}$
$\displaystyle{y = \tan^3 \sqrt{x}}$
$\displaystyle{y = \sqrt{x^3+6x}}$
$\displaystyle{y = \sin(\sec x)}$
$\displaystyle{y = x^3 \sqrt{x+1}}$
$\displaystyle{f\,(x) = \tan^2 \sqrt{x} - \sec^2 \sqrt{x}}$
$\displaystyle{g(x) = (\ln \sin^2 x)^3}$
$\displaystyle{h(x) = \ln \left[ \frac{x^2 e^{x^2}}{(x^2-2)^2} \right]}$
$\displaystyle{y = \cos( \tan^3 4x)}$
$\displaystyle{y = \frac{x^3-3x}{x-1}}$
$\displaystyle{y = (x-4)^{2/3} (x-1)^{1/3}}$
$\displaystyle{y = \ln \frac{e^{4x+5}}{\sqrt{x^3-e^x}}}$
$\displaystyle{h(x) = \frac{\cot^2 3x}{\cos^2 3x}}$
$\displaystyle{y = \sqrt{\ln x - \sqrt{e^x}}}$
$\displaystyle{f\,(x) = (x+4)^{1/3} (4-2x)^{2/3}}$
Differentiate the following. Simplify, as appropriate.
$\displaystyle{y = x(2-x)^{1/3}}$
$\displaystyle{y = \csc^2 x \cot^3 x}$
$\displaystyle{y = \frac{\sqrt{25-x^2}}{x}}$
$\displaystyle{y = \frac{\tan 3x}{\cot 3x}}$
$\displaystyle{y = \frac{(x-2)^{2/3}}{2x^{1/3}}}$
$\displaystyle{y = \frac{x}{\sqrt{4-x^2}}}$
$\displaystyle{y = \sin(\cot^2 3x^2)}$
$\displaystyle{y = \frac{x^2+4}{x-2}}$
$\displaystyle{y=3x^5 -2x +5 =\frac{1}{x^4}}$
$\displaystyle{f(x) = (4\sqrt{x} - 3)(\tan x)}$
$\displaystyle{f(x) = \frac{\ln x}{x^2-4}}$
$\displaystyle{y = \sin(\cos \sqrt{x+4} )}$
$\displaystyle{f(x) = 5(8-x^3)^{2/3} - \csc \sqrt{3x^2}}$
$\displaystyle{y = \frac{3^{3x^2} (\cot x)}{\sec x^2}}$
$\displaystyle{f(x) = \left( \frac{4x^6 - 3 \sqrt[4]{x}}{\ln(x^2-4)} \right)^8}$
$\displaystyle{y = \ln (\sin x)^{\cos x}}$
$\displaystyle{y = (\ln x)^{\cos x}}$
$\displaystyle{f(x) = \ln(\cos x)^x + x^{\cos x}}$
$\displaystyle{y = \frac{x^{2/3}}{x-4}}$
$\displaystyle{y = \frac{4x^3}{x^3+1}}$
$\displaystyle{y = x^{2/3} (1-2x)^{1/3}}$
$\displaystyle{y = x^{4/3} (x-2)^{1/3}}$
$\displaystyle{f(x) = \cos^2 (3x) \csc^2 (3x)}$
$\displaystyle{f(x) = \sqrt{x} \sin (\cos \sqrt{x})}$
Differentiate the following. Simplify, as appropriate.
$\displaystyle{y = \textrm{Arctan}^2 (\sqrt{x})}$
$\displaystyle{y = x \textrm{ Arcsin } x^3}$
$\displaystyle{y = \textrm{Arctan }(\sin x)}$
$\displaystyle{y = \textrm{Arcsin}^3 (3x^3)}$
Differentiate the following, but do not simplify.
$\displaystyle{y = \textrm{Arcsin } \sqrt{x} + \textrm{ Arctan } x^2 - \ln x^x}$
$\displaystyle{y = \sin [ \cos ( e^{\sqrt{x}} ) ]}$
$\displaystyle{f(x) = \frac{\sqrt{2x}}{3} + \frac{4}{\sqrt{x}} - \pi x}$
$\displaystyle{\sec^3(3 \sqrt[3]{x})}$
Differentiate the following. Simplify, as appropriate.
$\displaystyle{y = x^{1/3}(4-x)^{2/3}}$
$\displaystyle{y = \ln \sqrt{\frac{e^{3x} (x-4)}{x^3}}}$
$\displaystyle{f(x) = \textrm{Arctan } x - \textrm{Arctan } \left( \frac{1}{x} \right)}$
$\displaystyle{f(x) = \ln(\cos x)}$
$\displaystyle{y=\frac{\sin^2(4x) \sec^2(4x)}{\tan^2(4x)\csc^2(4x)}}$
$\displaystyle{y = (1-x)^{1/3}(x+2)^{5/3}}$
$\displaystyle{y = \frac{x}{\sqrt{16-x^2}}}$
$\displaystyle{y = x(x-1)^3}$
$\displaystyle{y = x^{1/3}(2-x)^{2/3}}$
$\displaystyle{y = \textrm{Arcsin } \sqrt{x}}$
$\displaystyle{\ln \left( \frac{x \sec 3x}{e^{\tan x^2}} \right)}$
$\displaystyle{(x^2+4)^{\cos 5x}}$
$\displaystyle{(\ln x + e^x)^{\ln(\ln(x^2+1))}}$
$\displaystyle{y = \left( \frac{x+1}{2x+3} \right)}$
$\displaystyle{y = (x+1)^{1/3} (x-1)^{2/3}}$
$\displaystyle{y = \cot^2 (x^3) \sec^2 (x^3)}$
$\displaystyle{y = \frac{e^x - e^{-x}}{2}}$
$\displaystyle{y = \ln \sqrt{\frac{x-1}{x+1}}}$
$\displaystyle{y = \sin^3 x^3}$
$y = (x^2+\frac{1}{x})^5$
$\displaystyle{y = \cos x^2 - \cos(2x) + 2\cos^2 x}$
$\displaystyle{y = \left[ \ln e^{-x} \right]^2 + \ln x^2 + \ln \left( \frac{e^x}{\sqrt{x}} \right)}$
$\displaystyle{y = \ln \left( \frac{x e^{x^2-4}}{\tan^2 x} \right)}$
$\displaystyle{y = (\sin x^2 )^{\ln \cos x}}$
$\displaystyle{y = \ln (4x^3)^2}$
Find the tangent line to the graph of $f\,(x) = (x^2-\frac{1}{x})(x^2+1)$ at $x=1$.
Where are there horizontal tangent lines to the graph of $y=\sqrt[3]{2x^3-24x}$?
Find the normal line to the graph of $f\,(x) = x^2(3x+4)^{2/3}$ at $x=-1$.
Find the equation of the tangent line to $y=-\sqrt{25-x^2}$ at $x=4$.
Suppose $\displaystyle{f\,(x) = \textrm{Arctan } x + \textrm{Arctan } 1/x}$