Tolong dibantu ya tugas rumah saya yg ad pd gmbar d bawah ini

f(x) = [tex]\frac{x^{2}+10x+24}{x+2}[/tex]
Jika u = [tex]x^2[/tex] + 10x + 24
dan v = x + 2
Maka, ([tex]\frac{u}{v}[/tex])' = [tex]\frac{u'v-uv'}{v^{2}}[/tex]
u' = 2x + 10
v' = 1
[tex]\frac{u'v-uv'}{v^{2}}[/tex] = [tex]\frac{(2x+10)(x+2)-(x^{2}+10x+24)(1)}{(x+2)^{2}}[/tex] = [tex]\frac{(2x^{2}+4x+10x+20)-(x^{2}+10x+24)}{(x^{2}+4x+4)}[/tex] = [tex]\frac{2x^{2}+14x+20-x^{2}-10x-24}{x^{2}+4x+4}[/tex] = [tex]\frac{x^{2}+4x-4}{x^{2}+4x+4}[/tex]
Jadi, f'(x) = [tex]\frac{x^{2}+4x-4}{x^{2}+4x+4}[/tex]
f'(1) = [tex]\frac{1^{2}+4(1)-4}{1^{2}+4(1)+4}[/tex] = [tex]\frac{1+4-4}{1+4+4}[/tex] = [tex]\frac{1}{9}[/tex]