Dibantuin yah kakak yg pintar,cara kerjanya dituliskan yah. tentukan f'(2) dan f'(-1) dari fungsi be.... Question from @andiaini

October 2018 1 13 Report

Dibantuin yah kakak yg pintar,cara kerjanya dituliskan yah.
tentukan f'(2) dan f'(-1) dari fungsi berikut.

a. f(x)= 3akar (3x^2+2)
b. f(x)= (x^2-2x)^-3/5

izharulhaq78 $A.$
$f(x)=3\sqrt{3x^2+2}\\f(x)=3(3x^2+2)^\frac{1}{2}\\\\f'(x)=3\times6x\times\frac{1}{2}(3x^2+2)^{\frac{1}{2}-1}\\f'(x)=3\times3x(3x^2+2)^{-\frac{1}{2}}\\f'(x)=9x(3x^2+2)^{-\frac{1}{2}}\\f'(x)=9x\frac{1}{\sqrt{3x^2+2}}\\\\f'(2)=9(2)\frac{1}{\sqrt{3(2)^2+2}}\\f'(2)=18\frac{1}{\sqrt{3(4)+2}}\\f'(2)=18\frac{1}{\sqrt{12+2}}\\f'(2)=18\frac{1}{\sqrt{14}}\\f'(2)=\frac{18}{\sqrt{14}}\\f'(2)=\frac{18}{14}\sqrt{14}\\\boxed{f'(2)=\frac{9}{7}\sqrt{14}}\\\\f'(-1)=9(-1)\frac{1}{\sqrt{3(-1)^2+2}}$
$f'(-1)=-9\frac{1}{\sqrt{3(1)+2}}\\f'(-1)=-9\frac{1}{\sqrt{3+2}}\\f'(-1)=-9\frac{1}{\sqrt{5}}\\f'(-1)=\frac{-9}{\sqrt{5}}\\\boxed{f'(-1)=\frac{-9}{5}\sqrt{5}}$

$B.$
$f(x)=(x^2-2x)^{-\frac{3}{5}}\\\\f'(x)=-\frac{3}{5}(x^2-2x)^{-\frac{3}{5}-1}\times(2x-2)\\f'(x)=-\frac{3}{5}(2x-2)(x^2-2x)^{-\frac{8}{5}}\\f'(x)=-\frac{3}{5}(2x-2)\frac{1}{\sqrt[5]{(x^2-2x)^8}}\\f'(x)=-3(2x-2)\frac{1}{5(x^2-2x)\sqrt[5]{(x^2-2x)^3}}\\f'(x)=\frac{-3(2x-2)}{5(x^2-2x)\sqrt[5]{(x^2-2x)^3}}\\\\f'(2)=\frac{-3(2(2)-2)}{5(2^2-2(2))\sqrt[5]{(2^2-2(2))^3}}\\f'(2)=\frac{-3(4-2)}{5(4-4)\sqrt[5]{(4-4)^3}}\\f'(2)=\frac{-3(2)}{5(0)\sqrt[5]{(0)^3}}\\\boxed{f'(2)=\infty}$
$f'(-1)=\frac{-3(2(-1)-2)}{5((-1)^2-2(-1))\sqrt[5]{((-1)^2-2(-1))^3}}\\f'(-1)=\frac{-3(-2-2)}{5(1+2)\sqrt[5]{(1+2)^3}}\\f'(-1)=\frac{-3(-4)}{5(3)\sqrt[5]{(3)^3}}\\f'(-1)=\frac{-1(-4)}{5\sqrt[5]{27}}\\f'(-1)=\frac{4}{5\sqrt[5]{27}}\\f'(-1)=\frac{4}{5\times27}\sqrt[5]{27}\\\boxed{f'(-1)=\frac{4}{135}\sqrt[5]{27}}$

Maaf kalau salah

0 votes Thanks 1

andiaini makasih kak,bantu kerjainnomor 2 yah

izharulhaq78 Yang no. 2 itu pangkat -3/5?

andiaini iya kak pangkat -3/5

andiaini boleh kerjain tugas saya 1 lagi?