differentiation

[Lippi86]

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..

 

[soroban]

Hello, Lippi86!

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

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

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

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

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

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


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