Hallo,
1.Aufgabe:
y= 3*(10-x)+8*√(x²+25)
y1= 3*(10-x) =30-3x
y1'= -3
y2= 8*√(x²+25) = 8 *(x^2+25)^(1/2)
y2'= 8 * 1/2 (x^2+25)^(-1/2) *2x ->Kettenregel
y2'= 8x (x^2+25)^(-1/2)
y' = y1' +y2'
y'= -3 +8x (x^2+25)^(-1/2)
\( \frac{d}{d x}(3(10-x)+8 \sqrt{x^{2}+25})=\frac{8 x}{\sqrt{x^{2}+25}}-3 \)