Svar 1.2:1
Förberedande kurs i matematik 2
(Skillnad mellan versioner)
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%"| <math>\cos^2x-\sin^2x=\cos2x</math> |b) |width="33%"| <math>2x\ln x+ x</math> |c) |width="33%"| <math>\displaystyle\frac{x^2+2x-1}{(x+1)...) |
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%"| <math>\cos^2x-\sin^2x=\cos2x</math> |b) |width="33%"| <math>2x\ln x+ x</math> |c) |width="33%"| <math>\displaystyle\frac{x^2+2x-1}{(x+1)...) |
Nuvarande version
| a) | \displaystyle \cos^2x-\sin^2x=\cos2x | b) | \displaystyle 2x\ln x+ x | c) | \displaystyle \displaystyle\frac{x^2+2x-1}{(x+1)^2}=1-\frac{2}{(x+1)^2} |
| d) | \displaystyle \displaystyle\frac{\cos x}{x}-\frac{\sin x}{x^2} | e) | \displaystyle \displaystyle\frac{1}{\ln x}-\frac{1}{(\ln x)^2} | f) | \displaystyle \displaystyle \frac{\ln x + 1}{\sin x}-\frac{x\ln x \cos x}{\sin^2x} |
