Partial Differentiation
2017/08/03 00:11
瀏覽124
迴響0
推薦0
引用0
標題:
發問:
Find dw/dt if w=x^2+y^2-z^2 , x=ln t , y=t^3 , z=sin t^2
最佳解答:
if w=x^2+y^2-z^2 , x=ln t , y=t^3 , z=sin t^2 w = (ln t)^2 + (t^3)^2 - [sin (t^2)]^2 dw/dt = [ (ln t)^2 ]' + [ (t^3)^2 ]' - { [sin (t^2)]^2 }' dw/dt = 2 (ln t) (1/t) + 6t^5 - 2 sin(t^2) cos(t^2) ? 2t dw/dt = (2/t) (ln t) + 6t^5 - 4t sin(t^2) cos(t^2) . . . dw/dt = (2/t) (ln t) + 6t^5 - 2t sin(2t^2)
其他解答:
Apologize for poor IT: 圖片參考：https://s.yimg.com/rk/HA08385518/o/12246635.jpg
此文章來自奇摩知識+如有不便請留言告知
Partial Differentiation發問:
Find dw/dt if w=x^2+y^2-z^2 , x=ln t , y=t^3 , z=sin t^2
最佳解答:
if w=x^2+y^2-z^2 , x=ln t , y=t^3 , z=sin t^2 w = (ln t)^2 + (t^3)^2 - [sin (t^2)]^2 dw/dt = [ (ln t)^2 ]' + [ (t^3)^2 ]' - { [sin (t^2)]^2 }' dw/dt = 2 (ln t) (1/t) + 6t^5 - 2 sin(t^2) cos(t^2) ? 2t dw/dt = (2/t) (ln t) + 6t^5 - 4t sin(t^2) cos(t^2) . . . dw/dt = (2/t) (ln t) + 6t^5 - 2t sin(2t^2)
其他解答:
Apologize for poor IT: 圖片參考：https://s.yimg.com/rk/HA08385518/o/12246635.jpg
你可能會有興趣的文章：
限會員,要發表迴響,請先登入
