I need help with this question please a If fx ecosx where 0

I need help with this question, please.

(a) If f(x) = e^cosx where 0 lessthanorequalto x lessthanorequalto pi/2, what is (f^-1)\' (squareroot e)? (b) If g(x) = x^2 + ln x where x > 0, what is (f^-1)\' (1)?

Solution

2)

when f has inverse, then (f(f^-1(x)))=x

differentiate with respect to x

derivative rule: (f(g(x)))\'=(f \'(g(x)))*(g\'(x))

=>(f \'(f^-1(x)))*(f^-1(x))\'=1

=>(f^-1(x))\'=1_{/(f \'(f^-1(x)))}

(a)

e^cosx=e

=>e^cosx=e^1/2

=>cosx=1/2

=>x=_/3

so f(_/3)=e

=>f^-1(e)=(_/3)

f(x)=e^cosx

f \'(x)=-sinx*e^cosx

f \'(_/3)=-sin(_/3)*e^cos(_/3)

f \'(_/3)=-(3 _/2)e

f \'(_/3)=-((3e))_/2

(f^-1(e))\'=1_{/(f \'(f^-1(e)))}

(f^-1(e))\'=1_{/(f \'(/3))}

(f^-1(e))\'=-2_/(3e)

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(b)

x²+ln(x)=1

=>x =1

so ,g(1)=1

=>g^-1(1)=1

g(x)=x²+ln(x)

=>g\'(x)=2x+(1_/x)

=>g\'(1)=(2*1)+(1_/1)

=>g\'(1)=3

(g^-1(1))\'=1_{/(g\'(g^-1(1)))}

=>(g^-1(1))\'=1_/(g\'(1))

=>(g^-1(1))\'=1_/3