How to find derivate of this function y=3^(x^2-x+1) ?
回答 (6)
2016-03-17 9:22 am
y = 3^(x^2 - x + 1)
ln(y) = ln[3^(x^2 - x + 1)]
ln(y) = (x^2 - x + 1) ln(3)
d ln y / dx = d [(x^2 - x + 1) ln(3)] / dx
(1/y) (dy/dx) = (2x - 1) ln(3)
dy/dx = y (2x - 1) ln(3)
dy/dx = [3^(x^2 - x + 1)] (2x - 1) ln(3)
ln(y) = ln[3^(x^2 - x + 1)]
ln(y) = (x^2 - x + 1) ln(3)
d ln y / dx = d [(x^2 - x + 1) ln(3)] / dx
(1/y) (dy/dx) = (2x - 1) ln(3)
dy/dx = y (2x - 1) ln(3)
dy/dx = [3^(x^2 - x + 1)] (2x - 1) ln(3)
2016-03-17 1:25 pm
y = 3^(x^2-x+1)
ln y = (x^2-x+1) ln 3
Take derivative with respect to x from both sides
(1/y) dy/dx = (2x-1) ln 3
dy/dx = y (2x-1) ln 3
dy/dx = 3^(x^2-x+1) (2x-1) ln 3
ln y = (x^2-x+1) ln 3
Take derivative with respect to x from both sides
(1/y) dy/dx = (2x-1) ln 3
dy/dx = y (2x-1) ln 3
dy/dx = 3^(x^2-x+1) (2x-1) ln 3
2016-03-17 11:12 am
log y = [ x² - x + 1 ] log 3
(1/y) dy/dx = [2 log 3] x - log 3
dy/dx = y [2 log 3] x - y log 3
dy/dx = 3^(x² - x + 1) [ (2 log 3) x - log 3 ]
(1/y) dy/dx = [2 log 3] x - log 3
dy/dx = y [2 log 3] x - y log 3
dy/dx = 3^(x² - x + 1) [ (2 log 3) x - log 3 ]
2016-03-17 9:17 am
y= 3^( x^2-x+1)
dy/dx = d/dx[ 3^(x^2-x+1)
= 3^(x^2-x+1) *log3 *d/dx(x^2-x+1) { d/dx a^x = a^x loga }
= 36(x^2-x+1) * log 3* [ 2x-1] answer
dy/dx = d/dx[ 3^(x^2-x+1)
= 3^(x^2-x+1) *log3 *d/dx(x^2-x+1) { d/dx a^x = a^x loga }
= 36(x^2-x+1) * log 3* [ 2x-1] answer
2016-03-17 8:51 am
Notice that d/dx(3ˣ) = ln(3) • 3ˣ
By chain rule, d/dx(3^ƒ(x)) = ln(3) • 3^(ƒ(x)) • ƒ'(x)
d/dx(3^(x² - x + 1)) = ln(3) • 3^(x² - x + 1) • (2x - 1)
By chain rule, d/dx(3^ƒ(x)) = ln(3) • 3^(ƒ(x)) • ƒ'(x)
d/dx(3^(x² - x + 1)) = ln(3) • 3^(x² - x + 1) • (2x - 1)
2016-03-17 8:51 am
Use the fact that d/dx a^x = a^x ln(a), along with chain rule to get dy/dx = 3^(x^2-x+1)*(2*x-1)*ln(3).
收錄日期: 2021-04-18 14:38:43
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