Question

Найти f '(1),если:

f(x)=((2x-1)^5)*((1+x)^4)

f(x)=((5x-4)^6)*sqrt(3x-2)

Answer 1

$\f(x)=(2x-1)^5(1+x)^4\ f'(x)=5(2x-1)^4cdot2(1+x)^4+(2x-1)^5cdot 4(1+x)^3\ f'(x)=(2x-1)^4(1+x)^3(5cdot2(1+x)+(2x-1)cdot4)\ f'(x)=(2x-1)^4(1+x)^3(10+10x+8x-4)\ f'(x)=(2x-1)^4(1+x)^3(18x+6)\ f'(x)=6(2x-1)^4(1+x)^3(3x+1)\ \ f'(x)=6(2x-1)^4(1+x)^3(3x+1)\ f'(1)=6(2cdot1-1)^4(1+1)^3(3cdot 1+1)\ f'(1)=6cdot1cdot 8cdot 4\ f'(1)=192$

 

 

$\f(x)=(5x-4)^6sqrt{3x-2}\ f'(x)=6(5x-4)^5cdot5sqrt{3x-2}+(5x-4)^6cdotfrac{1}{2sqrt{3x-2}}cdot 3\ f'(x)=3(5x-4)^5(2cdot5sqrt{3x-2}+frac{5x-4}{2sqrt{3x-2}})\ f'(x)=3(5x-4)^5(10sqrt{3x-2}+frac{(5x-4)(sqrt{3x-2})}{2(3x-2)})\ f'(x)=3(5x-4)^5sqrt{3x-2}(10+frac{5x-4}{6x-4})\ f'(x)=3(5x-4)^5sqrt{3x-2}(frac{10(6x-4)}{6x-4}+frac{5x-4}{6x-4})\ f'(x)=3(5x-4)^5sqrt{3x-2}(frac{60x-40+5x-4}{6x-4})\ f'(x)=3(5x-4)^5sqrt{3x-2}(frac{65x-44}{6x-4})\ \\ f'(1)=3(5cdot 1-4)^5sqrt{3cdot 1-2}(frac{65cdot 1-44}{6cdot 1 -4})\ f'(1)=3cdot1cdot 1cdot frac{21}{2}\ f'(1)=frac{63}{2}$