L Lippi86 New member Joined Feb 22, 2010 Messages 18 Feb 27, 2010 #1 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..
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..
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Feb 27, 2010 #2 Hello, Lippi86! It's a straight-forward differentiation . . . and an evaluation. Exactly where is your difficulty? \(\displaystyle f(x)\:=\:\ln\left(\sqrt{x} +x^2\right)\) \(\displaystyle \text{Calculate: }\: f'(1)\) Click to expand... \(\displaystyle \text{We have: }\;f(x) \;=\;\ln\left(x^{\frac{1}{2}} + x^2\right)\) \(\displaystyle \text{Then: }\;f'(x) \;=\;\frac{1}{x^{\frac{1}{2}} + x^2}\cdot\left(\tfrac{1}{2}x^{-\frac{1}{2}} + 2x\right) \;=\;\frac{\frac{1}{2x^{\frac{1}{2}}} + 2x}{x^{\frac{1}{2}} + x^2}\) \(\displaystyle \text{Multiply by }\frac{2x^{\frac{1}{2}}} {2x^{\frac{1}{2}}}}: \;\; f'(x) \;=\;\frac{1 + 4x^{\frac{3}{2}}} {2x + 2x^{\frac{5}{2}}}\) \(\displaystyle \text{Therefore: }\;f'(1) \;=\;\frac{1 + 4(1^{\frac{3}{2}})} {2(1) + 2(1^{\frac{5}{2}})} \;=\; \frac{1+4}{2+2} \;=\;\frac{5}{4}\)
Hello, Lippi86! It's a straight-forward differentiation . . . and an evaluation. Exactly where is your difficulty? \(\displaystyle f(x)\:=\:\ln\left(\sqrt{x} +x^2\right)\) \(\displaystyle \text{Calculate: }\: f'(1)\) Click to expand... \(\displaystyle \text{We have: }\;f(x) \;=\;\ln\left(x^{\frac{1}{2}} + x^2\right)\) \(\displaystyle \text{Then: }\;f'(x) \;=\;\frac{1}{x^{\frac{1}{2}} + x^2}\cdot\left(\tfrac{1}{2}x^{-\frac{1}{2}} + 2x\right) \;=\;\frac{\frac{1}{2x^{\frac{1}{2}}} + 2x}{x^{\frac{1}{2}} + x^2}\) \(\displaystyle \text{Multiply by }\frac{2x^{\frac{1}{2}}} {2x^{\frac{1}{2}}}}: \;\; f'(x) \;=\;\frac{1 + 4x^{\frac{3}{2}}} {2x + 2x^{\frac{5}{2}}}\) \(\displaystyle \text{Therefore: }\;f'(1) \;=\;\frac{1 + 4(1^{\frac{3}{2}})} {2(1) + 2(1^{\frac{5}{2}})} \;=\; \frac{1+4}{2+2} \;=\;\frac{5}{4}\)
