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f(x)=sec(x) what is f"(pi/4)? Explain procedure
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f(x)=sec(x) f'(x)=sec(x)*tan(x) f''(x)=d/dx(sec(x))*tan(x)+sec(x)*(d/dx)(tan(x)) f''(x)=sec(x)*tan(x)*tan(x)+sec(x)*sec^2(x) f''(x)=sec(x)*tan^2(x)+sec^3(x) f''(pi/4)=sec(pi/4)*tan^2(pi/4)+sec^3(pi/4) f''(pi/4)=sqrt(2)*(1)^2+(sqrt(2))^3 f''(pi/4)=sqrt(2)+2sqrt(2) f''(pi/4)=3*sqrt(2)
\[f(x)= (\cos(x))^{-1}\]
use chain rule
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