differentiation

Lippi86

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Feb 22, 2010
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18
f(x)=ln(?(x)+x2).

Calculate f´(1).

Answers would be with a fraction simåliefied as much as possible...

Is anyone have an idea får have I start with these?

thank you=)

have a nice weekend..
 
Hello, Lippi86!

It's a straight-forward differentiation . . . and an evaluation.
Exactly where is your difficulty?


f(x)=ln(x+x2)\displaystyle f(x)\:=\:\ln\left(\sqrt{x} +x^2\right)

Calculate: f(1)\displaystyle \text{Calculate: }\: f'(1)

We have:   f(x)  =  ln(x12+x2)\displaystyle \text{We have: }\;f(x) \;=\;\ln\left(x^{\frac{1}{2}} + x^2\right)

Then:   f(x)  =  1x12+x2(12x12+2x)  =  12x12+2xx12+x2\displaystyle \text{Then: }\;f'(x) \;=\;\frac{1}{x^{\frac{1}{2}} + x^2}\cdot\left(\tfrac{1}{2}x^{-\frac{1}{2}} + 2x\right) \;=\;\frac{\frac{1}{2x^{\frac{1}{2}}} + 2x}{x^{\frac{1}{2}} + x^2}

\(\displaystyle \text{Multiply by }\frac{2x^{\frac{1}{2}}} {2x^{\frac{1}{2}}}}: \;\; f'(x) \;=\;\frac{1 + 4x^{\frac{3}{2}}} {2x + 2x^{\frac{5}{2}}}\)


Therefore:   f(1)  =  1+4(132)2(1)+2(152)  =  1+42+2  =  54\displaystyle \text{Therefore: }\;f'(1) \;=\;\frac{1 + 4(1^{\frac{3}{2}})} {2(1) + 2(1^{\frac{5}{2}})} \;=\; \frac{1+4}{2+2} \;=\;\frac{5}{4}

 
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