f(x)=2 ln [cosx]
what is f''(x)=?
f(x)=2ln[cos(x)]
By chain rule:
f'(x) = 2 * (1 / cos(x)) * -sin(x)
= -2sin(x)/cos(x)
= -2tan(x)
Use the chain rule.
d/dx(ln(x))=1/x
d/dx(cos(x))=-sin(x)
so
f'(x) = 2*(1/cos(x))*(-sin(x))
f'(x)=-2*tan(x)
f´(x) = 2/cos(x) * (-sin x)
Comments
f(x)=2ln[cos(x)]
By chain rule:
f'(x) = 2 * (1 / cos(x)) * -sin(x)
= -2sin(x)/cos(x)
= -2tan(x)
Use the chain rule.
d/dx(ln(x))=1/x
d/dx(cos(x))=-sin(x)
so
f'(x) = 2*(1/cos(x))*(-sin(x))
f'(x)=-2*tan(x)
f´(x) = 2/cos(x) * (-sin x)