đạo hàm hàm số sau: x7-7x^8-3/4x^3

Đáp án:

$y' = \dfrac{-35x^8 + 4x^7 + 9}{4x^4}$

Giải thích các bước giải:

$y = \dfrac{x^7 - 7x^8 - 3}{4x^3}$

$TXĐ: D = \Bbb R \backslash\left\{0\right\}$

$\to y' = \dfrac{(x^7 - 7x^8 - 3)'.(4x^3) - (4x^3)'(x^7 - 7x^8 - 3)}{16x^6}$

$\to y' = \dfrac{(7x^6 - 56x^7).4x^3 - 12x^2.(x^7 - 7x^8 - 3)}{16x^6}$

$\to y' = \dfrac{28x^9 - 224x^{10} - 12x^9 + 84x^{10} + 36x^2}{16x^6}$

$\to y' = \dfrac{-140x^{10} + 16x^9 + 36x^2}{16x^6}$

$\to y' = \dfrac{-35x^8 + 4x^7 + 9}{4x^4}$

$y=\dfrac{x^7-7x^8-3}{4x^3}$

$y'=\dfrac{(x^7-7x^8-3)'.4x^3-(x^7-7x^8-3)(4x^3)' }{16x^6}$

$=\dfrac{(7x^6-56x^7).4x^3-12x^2(x^7-7x^8-3) }{16x^6}$

$=\dfrac{28x^9-224x^{10}-12x^9+84x^{10}+36x^2 }{16x^6}$

$=\dfrac{-140x^{10}+16x^9+36x^2}{16x^6}$

$=\dfrac{-35x^8+4x^7+9}{4x^4}$