derivative100 [71-80]

71. $\frac{d}{dx}arctan(2x+3)$

$\begin{aligned} &\frac{d}{dx}arctan(2x+3)\\ &=\frac{1}{1+(2x+3)^2}(2x+3)'\\ &=\frac{2}{4x^2+12x+10}=\frac{1}{2x^2+6x+5} \end{aligned}$

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72. $\frac{d}{dx}cot^4(2x)$

$\begin{aligned} &\frac{d}{dx}cot^4(2x)\\ &=4cot^3(2x)(cot(2x))'=4cot^3(2x)(-csc^2(2x))(2x)'\\ &=-8cot^3(2x)csc^2(2x) \end{aligned}$

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73. $\frac{d}{dx}\frac{x^2}{1+\frac{1}{x}}$

$\begin{aligned} &\frac{d}{dx} \frac{x^2}{1+\frac{1}{x}}=\frac{d}{dx} \frac{x^2}{\frac{x+1}{x}}=\frac{d}{dx} \frac{x^3}{x+1}\\ &=\frac{3x^2(x+1)-x^3}{(x+1)^2}=\frac{2x^3+3x^2}{(x+1)^2} \end{aligned}$

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74. $\frac{d}{dx}e^{x/(1+x^2)}$

$\begin{aligned} &\frac{d}{dx}e^{\frac{x}{1+x^2}}=e^{\frac{x}{1+x^2}} (\frac{x}{1+x^2})'\\ &=e^{\frac{x}{1+x^2}} \frac{(1+x^2)-x(2x)}{(1+x^2)^2}\\ &=e^{\frac{x}{1+x^2}} \frac{1-x^2}{(1+x^2)^2}\\ \end{aligned}$

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75. $\frac{d}{dx}(arcsinx)^3$

$\begin{aligned} &\frac{d}{dx}(arcsinx)^3=3(arcsinx)^2\frac{1}{\sqrt{1-x^2}}\\ &=\frac{3(sin^{-1}x)^2}{\sqrt{1-x^2}} \end{aligned}$

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76. $\frac{d}{dx}\frac{1}{2} sec^2(x) – ln(secx)$

$\begin{aligned} &\frac{d}{dx}\frac{1}{2} sec^2(x) – ln(secx)\\ &=sec(x)sec(x)tan(x)-\frac{1}{sec(x)}sec(x)tan(x)\\ &=sec^2(x)tan(x)-tan(x)=(sec^2x-1)tanx\\ &=tan^3x \end{aligned}$

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77. $\frac{d}{dx}ln(ln(lnx)))$

$\begin{aligned} &\frac{d}{dx}ln(ln(lnx)))\\ &=\frac{1}{ln(ln(x))}(ln(ln(x)))'=\frac{1}{ln(ln(x))} \frac{1}{ln(x)}(ln(x))'\\ &=\frac{1}{xln(x)ln(ln(x))} \end{aligned}$

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78. $\frac{d}{dx}\pi^3$

$\begin{aligned} &\frac{d}{dx}\pi^3=0\\ \end{aligned}$

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79. $\frac{d}{dx}ln[x+\sqrt{1+x^2}]$

$\begin{aligned} &\frac{d}{dx}ln[x+\sqrt{1+x^2}]\\ &=\frac{1}{x+\sqrt{1+x^2}}(1+\frac{2x}{2\sqrt{1+x^2}})\\ &=\frac{1}{x+\sqrt{1+x^2}} \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\\ &=\frac{1}{\sqrt{1+x^2}} \end{aligned}$

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80. $\frac{d}{dx}arcsinh(x)$

$\begin{aligned} &\frac{d}{dx}arcsinh(x)\\ &=\frac{1}{\sqrt{1+x^2}}\\ & y=arcsinh(x), x=sinh(y)=\frac{e^y-e^{-y}}{2}\\ & 2x=e^y-e^{-y} \\ & 2dx = e^ydy+e^{-y}dy=dy(e^y+e^{-y})\\ & \frac{dy}{dx}=\frac{2}{e^y+e^{-y}}=\frac{1}{coshy}\\ &(cosh^2x-sinh^2x = 1, coshx=\sqrt{1+sinh^2x})\\ &\frac{1}{coshy}=\frac{1}{\sqrt{1+sinh^2y}}=\frac{1}{\sqrt{1+(\frac{e^y-e^{-y}}{2})^2}}\\ &=\frac{1}{\sqrt{1+(\frac{2x}{2})^2}}=\frac{1}{\sqrt{1+x^2}}\\ &\\ & Alt.\\ & \frac{dy}{dx}=\frac{2}{e^y+e^{-y}}\\ & x^2=\frac{e^{2y}+e^{-2y}-2}{4}=\frac{e^{2y}+e^{-2y}+2}{4}-1\\ & x^2+1 = (\frac{e^y+e^{-y}}{2})^2\\ & \frac{dy}{dx}=\frac{1}{\sqrt{1+x^2}} \end{aligned}$

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