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Calculus implicit differentiation help please????
I need help with trig implicit differentiation. I don't know how to do the problems, if I could get help on one problem then I'll know how to do the rest. Please. 3x+1= sec y cubed
3x+1= Sec y³
2 Answers
- cidyahLv 78 years ago
3x+1 = sec y^3
differentiate both sides with respect to x
3 = sec(y^3) tan(y^3) d/dx (y^3)
3 = sec (y^3) tan(y^3) (3y^2) dy/dx
3 = 3 y^2 sec(y^3) tan(y^3) dy/dx
dy/dx = 1 / (y^2 sec(y^3) tan(y^3) )
dy/dx = cos(y^3) cot(y^3) / y^2
- ?Lv 78 years ago
d/dx sec(y^3) = sec(y^3)*tan(y^3)*(3y^2)*y'
d/dx 3x+1 = 3 so
3=sec(y^3)*tan(y^3)*(3y^2)*y' and
y' = 3/sec(y^3)*tan(y^3)*(3y^2) = 3cos(y^3)cot(y^3)/3(y^2) = (y^(-2)*(cos(y^3)cot(y^3))
