Mohon dibantu :)
kalkulus ,,,,!
1.x^2-2x+5/x^2+2x-3
kalkulus ,,,,!
1.x^2-2x+5/x^2+2x-3
izharulhaq78
![\displaystyle f(x)=\frac{x^2-2x+5}{x^2+2x-3}\\\\u=x^2-2x+5\\u'=2x-2\\v=x^2+2x-3\\v'=2x+2\\\\f'(x)=\frac{u'v-uv'}{v^2}\\f'(x)=\frac{(2x-2)(x^2+2x-3)-(x^2-2x+5)(2x+2)}{(x^2+2x-3)^2}\\f'(x)=\frac{2x^3+2x^2-10x+6-(2x^3-2x^2+6x+10)}{(x^2+3x-x-3)^2}\\f'(x)=\frac{2x^3+2x^2-10x+6-2x^3+2x^2-6x-10}{[(x+3)(x-1)]^2}\\\boxed{\boxed{f'(x)=\frac{4x^2-16x-4}{[(x+3)(x-1)]^2}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle+f%28x%29%3D%5Cfrac%7Bx%5E2-2x%2B5%7D%7Bx%5E2%2B2x-3%7D%5C%5C%5C%5Cu%3Dx%5E2-2x%2B5%5C%5Cu%27%3D2x-2%5C%5Cv%3Dx%5E2%2B2x-3%5C%5Cv%27%3D2x%2B2%5C%5C%5C%5Cf%27%28x%29%3D%5Cfrac%7Bu%27v-uv%27%7D%7Bv%5E2%7D%5C%5Cf%27%28x%29%3D%5Cfrac%7B%282x-2%29%28x%5E2%2B2x-3%29-%28x%5E2-2x%2B5%29%282x%2B2%29%7D%7B%28x%5E2%2B2x-3%29%5E2%7D%5C%5Cf%27%28x%29%3D%5Cfrac%7B2x%5E3%2B2x%5E2-10x%2B6-%282x%5E3-2x%5E2%2B6x%2B10%29%7D%7B%28x%5E2%2B3x-x-3%29%5E2%7D%5C%5Cf%27%28x%29%3D%5Cfrac%7B2x%5E3%2B2x%5E2-10x%2B6-2x%5E3%2B2x%5E2-6x-10%7D%7B%5B%28x%2B3%29%28x-1%29%5D%5E2%7D%5C%5C%5Cboxed%7B%5Cboxed%7Bf%27%28x%29%3D%5Cfrac%7B4x%5E2-16x-4%7D%7B%5B%28x%2B3%29%28x-1%29%5D%5E2%7D%7D%7D "\displaystyle f(x)=\frac{x^2-2x+5}{x^2+2x-3}\\\\u=x^2-2x+5\\u'=2x-2\\v=x^2+2x-3\\v'=2x+2\\\\f'(x)=\frac{u'v-uv'}{v^2}\\f'(x)=\frac{(2x-2)(x^2+2x-3)-(x^2-2x+5)(2x+2)}{(x^2+2x-3)^2}\\f'(x)=\frac{2x^3+2x^2-10x+6-(2x^3-2x^2+6x+10)}{(x^2+3x-x-3)^2}\\f'(x)=\frac{2x^3+2x^2-10x+6-2x^3+2x^2-6x-10}{[(x+3)(x-1)]^2}\\\boxed{\boxed{f'(x)=\frac{4x^2-16x-4}{[(x+3)(x-1)]^2}}}")
0 votes
Thanks 1
tmultidona
terimakasih pak atas pelajarannya...semoga bisa dipahami.. :-)
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