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DomashHunter
@DomashHunter
July 2022
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Помогите решить производные сложной функции:
1) y= x^3+5tg2x
2) y= ln(1- cos^2 3x)
3) y=√^3 (4x-1)/(4x+1)
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Производные сложной функции:1) y= x^3+5tg2x
y⁽¹⁾= 3x^2+12(1/cos²2x)·2=3x^2+24(1/cos²2x)
2) y= ln(1- cos^2 3x)
y⁽¹⁾=[1/(1- cos² 3x)]·(-2 cos 3x)(-sin 3x)(3)=(6 cos 3x)(sin 3x)/(1- cos² 3x)
3) y=√^3 (4x-1)/(4x+1)
со степенью не совсем понятно ...какая...
y=[∛ (4x-1)]/(4x+1) или y=∛ [(4x-1)/(4x+1)] или еще как...
y=[∛ (4x-1)]/(4x+1)=(4x-1)^(1/3)/(4x+1)
y⁽¹⁾=[(1/3)(4x-1)^(-2/3) ·(4x+1)·4-4(4x-1)^(1/3)]/(4x+1)²=
=[4(4x+1)-12(4x-1)]/[(4x-1)^(2/3)(4x+1)²]=[-32x+16]/[(4x-1)^(2/3)(4x+1)²]
y=∛ [(4x-1)/(4x+1)]=[(4x-1)/(4x+1)]^(1/3)
y⁽¹⁾=(1/3)[(4x-1)/(4x+1)]^(-2/3) ·[4(4x+1)-4(4x+1)]/(4x+1)²=
=(8/3)/[(4x-1)^(2/3) (4x+1)^(4/3)]
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Verified answer
Производные сложной функции:1) y= x^3+5tg2xy⁽¹⁾= 3x^2+12(1/cos²2x)·2=3x^2+24(1/cos²2x)
2) y= ln(1- cos^2 3x)
y⁽¹⁾=[1/(1- cos² 3x)]·(-2 cos 3x)(-sin 3x)(3)=(6 cos 3x)(sin 3x)/(1- cos² 3x)
3) y=√^3 (4x-1)/(4x+1)
со степенью не совсем понятно ...какая...
y=[∛ (4x-1)]/(4x+1) или y=∛ [(4x-1)/(4x+1)] или еще как...
y=[∛ (4x-1)]/(4x+1)=(4x-1)^(1/3)/(4x+1)
y⁽¹⁾=[(1/3)(4x-1)^(-2/3) ·(4x+1)·4-4(4x-1)^(1/3)]/(4x+1)²=
=[4(4x+1)-12(4x-1)]/[(4x-1)^(2/3)(4x+1)²]=[-32x+16]/[(4x-1)^(2/3)(4x+1)²]
y=∛ [(4x-1)/(4x+1)]=[(4x-1)/(4x+1)]^(1/3)
y⁽¹⁾=(1/3)[(4x-1)/(4x+1)]^(-2/3) ·[4(4x+1)-4(4x+1)]/(4x+1)²=
=(8/3)/[(4x-1)^(2/3) (4x+1)^(4/3)]