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1. 

 

If \(f(x)=x^2\ \text{and} \ g(x)=\ln(x),\ \text{compute}\ f'(1) + g'(1)\)

 

2. \(\int_0^4\frac{dx}{\sqrt{|x-2|}}\)

Guest Jun 5, 2018
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\(1. \text{If}\ f(x) = x^2\ \text{and}\ g(x)=\ln(x), \text{compute } f'(1) + g'(1)\)

 

\(\text{We compute}\ f'(x) = 2x\ \text{and}\ g'(x)=\frac1x,\ \text{so plugging in 1 for both gives us}\ \boxed3. \)

 

\(2. \int_0^4\frac{dx}{\sqrt{|x-2|}}\)

 

\(\text{Since the function } \sqrt{|x-2|}\text{ discontinues at } x=2,\\ \text{we can split the integral into two parts and compute separately.}\)

 

\(\int_0^2\frac{dx}{\sqrt{|x-2|}}=\int_0^2\frac{dx}{\sqrt{2-x}}=-2\sqrt{2-x}|_0^2=2\sqrt2\)

\(\int_2^4\frac{dx}{\sqrt{|x-2|}}=\int_2^4\frac{dx}{\sqrt{x-2}}=2\sqrt{x-2}|_4^2=2\sqrt2\)

 

\(2\sqrt2+2\sqrt2=\boxed{4\sqrt2}\)

 

laughlaughlaugh

GYanggg  Jun 5, 2018
edited by GYanggg  Jun 5, 2018

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