Найти f '(1),если: f(x)=((2x-1)^5)*((1+x)^4) f(x)=((5x-4)^6)*sqrt(3x-2)

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

$\\f(x)=(5x-4)^6\sqrt{3x-2}\\ f'(x)=6(5x-4)^5\cdot5\sqrt{3x-2}+(5x-4)^6\cdot\frac{1}{2\sqrt{3x-2}}\cdot 3\\ f'(x)=3(5x-4)^5(2\cdot5\sqrt{3x-2}+\frac{5x-4}{2\sqrt{3x-2}})\\ f'(x)=3(5x-4)^5(10\sqrt{3x-2}+\frac{(5x-4)(\sqrt{3x-2})}{2(3x-2)})\\ f'(x)=3(5x-4)^5\sqrt{3x-2}(10+\frac{5x-4}{6x-4})\\ f'(x)=3(5x-4)^5\sqrt{3x-2}(\frac{10(6x-4)}{6x-4}+\frac{5x-4}{6x-4})\\ f'(x)=3(5x-4)^5\sqrt{3x-2}(\frac{60x-40+5x-4}{6x-4})\\ f'(x)=3(5x-4)^5\sqrt{3x-2}(\frac{65x-44}{6x-4})\\ \\\\ f'(1)=3(5\cdot 1-4)^5\sqrt{3\cdot 1-2}(\frac{65\cdot 1-44}{6\cdot 1 -4})\\ \ f'(1)=3\cdot1\cdot 1\cdot \frac{21}{2}\\ f'(1)=\frac{63}{2}$