What Is written 2

Calculus Level 4

Let f ( x ) f(x) be a real valued bijective function satisfying f ′ ( x ) = sin ⁡ 2 ( sin ⁡ ( x + 1 ) ) f'(x) = \sin^2 (\sin (x+1)) and f ( 0 ) = 3 f(0) = 3 . Find the value of ( f − 1 ) ′ ′ ( 3 ) \left( f^{-1}\right)'' (3) .

Clarifications :

  • f ′ ( x ) f'(x) denotes the first derivative of f ( x ) f(x) with respect to x x .

  • f ′ ′ ( x ) f''(x) denotes the second derivative of f ( x ) f(x) with respect to x x .

  • f − 1 ( x ) f^{-1} (x) denotes the inverse function of f ( x ) f(x) .

− 2 sin ⁡ cos ⁡ 1 sin ⁡ 2 1 sin ⁡ 5 cos ⁡ 1 -\frac { 2\sin { \cos { 1 } \sin ^{ 2 }{ 1 } } }{ \sin ^{ 5 }{ \cos { 1 } } } − sin ⁡ 2 sin ⁡ 1 cos ⁡ 2 cos ⁡ 1 -\frac { \sin ^{ 2 }{ \sin { 1 } } }{ \cos ^{ 2 }{ \cos { 1 } } } − 2 sin ⁡ cos ⁡ 1 sin ⁡ 1 sin ⁡ 5 cos ⁡ 1 -\frac { 2\sin { \cos { 1 } \sin { 1 } } }{ \sin ^{ 5 }{ \cos { 1 } } } − 2 sin ⁡ cos ⁡ 1 sin ⁡ 1 sin ⁡ 6 cos ⁡ 1 -\frac { 2\sin { \cos { 1 } \sin { 1 } } }{ \sin ^{ 6 }{ \cos { 1 } } } − 2 sin ⁡ cos ⁡ 1 sin ⁡ 2 1 sin ⁡ 6 cos ⁡ 1 -\frac { 2\sin { \cos { 1 } \sin ^{ 2 }{ 1 } } }{ \sin ^{ 6 }{ \cos { 1 } } } − 2 cos ⁡ 1 cos ⁡ sin ⁡ 1 sin ⁡ 6 sin ⁡ 1 -\frac { 2\cos { 1 } \cos { \sin { 1 } } }{ \sin ^{ 6 }{ \sin { 1 } } } − cos ⁡ 2 cos ⁡ 1 sin ⁡ 2 sin ⁡ 1 -\frac { \cos ^{ 2 }{ \cos { 1 } } }{ \sin ^{ 2 }{ \sin { 1 } } } − 2 cos ⁡ 1 cos ⁡ sin ⁡ 1 sin ⁡ 5 sin ⁡ 1 -\frac { 2\cos { 1 } \cos { \sin { 1 } } }{ \sin ^{ 5 }{ \sin { 1 } } }

Rishabh Deep Singh by Rishabh Deep Singh , 23, India

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1 solution

Samuel Lo
Oct 1, 2018

Step 1:

f ′ ( x ) = ( sin ⁡ ( sin ⁡ ( x + 1 ) ) ) 2 f ′ ( 0 ) = ( sin ⁡ ( sin ⁡ ( 1 ) ) ) 2 \begin{aligned} f^{\prime} (x) &= (\sin(\sin(x+1)))^2 \\ f^{\prime} (0) &= (\sin(\sin(1)))^2 \\ \end{aligned}

Step 2:

f ′ ( x ) = ( sin ⁡ ( sin ⁡ ( x + 1 ) ) ) 2 f ′ ′ ( x ) = 2 sin ⁡ ( sin ⁡ ( x + 1 ) ) ⋅ cos ⁡ ( sin ⁡ ( x + 1 ) ) ⋅ cos ⁡ ( x + 1 ) f ′ ′ ( 0 ) = 2 sin ⁡ ( sin ⁡ ( 1 ) ) ⋅ cos ⁡ ( sin ⁡ ( 1 ) ) ⋅ cos ⁡ ( 1 ) \begin{aligned} f^{\prime} (x) &= (\sin(\sin(x+1)))^2 \\ f^{\prime \prime} (x) &= 2 \sin(\sin(x+1)) \cdot \cos(\sin(x+1)) \cdot \cos(x+1) \\ f^{\prime \prime} (0) &= 2 \sin(\sin(1)) \cdot \cos(\sin(1)) \cdot \cos(1) \\ \end{aligned}

Step 3:

Because f f is bijective and f ( 0 ) = 3 f(0)=3 , f − 1 ( 3 ) = 0 f^{-1} (3) = 0

Step 4:

f ( f − 1 ( x ) ) = x f ′ ( f − 1 ( x ) ) ⋅ ( f − 1 ) ′ ( x ) = 1 ( f − 1 ) ′ ( x ) = 1 f ′ ( f − 1 ( x ) ) \begin{aligned} f(f^{-1}(x)) &= x \\ f^{\prime}(f^{-1}(x)) \cdot (f^{-1})^{\prime}(x) &= 1 \\ (f^{-1})^{\prime}(x) &= \frac{1}{ f^{\prime}(f^{-1}(x)) } \\ \end{aligned}

Step 5:

( f − 1 ) ′ ( x ) = 1 f ′ ( f − 1 ( x ) ) ( f − 1 ) ′ ′ ( x ) = − 1 ( f ′ ( f − 1 ( x ) ) ) 2 ⋅ f ′ ′ ( f − 1 ( x ) ) ⋅ ( f − 1 ) ′ ( x ) ( f − 1 ) ′ ′ ( x ) = − 1 ( f ′ ( f − 1 ( x ) ) ) 2 ⋅ f ′ ′ ( f − 1 ( x ) ) ⋅ 1 f ′ ( f − 1 ( x ) ) ( f − 1 ) ′ ′ ( x ) = − f ′ ′ ( f − 1 ( x ) ) ( f ′ ( f − 1 ( x ) ) ) 3 ( f − 1 ) ′ ′ ( 3 ) = − f ′ ′ ( f − 1 ( 3 ) ) ( f ′ ( f − 1 ( 3 ) ) ) 3 = − f ′ ′ ( 0 ) ( f ′ ( 0 ) ) 3 = − 2 sin ⁡ ( sin ⁡ ( 1 ) ) ⋅ cos ⁡ ( sin ⁡ ( 1 ) ) ⋅ cos ⁡ ( 1 ) ( ( sin ⁡ ( sin ⁡ ( 1 ) ) ) 2 ) 3 = − 2 sin ⁡ ( sin ⁡ ( 1 ) ) ⋅ cos ⁡ ( sin ⁡ ( 1 ) ) ⋅ cos ⁡ ( 1 ) ( sin ⁡ ( sin ⁡ ( 1 ) ) ) 6 = − 2 cos ⁡ ( sin ⁡ ( 1 ) ) ⋅ cos ⁡ ( 1 ) ( sin ⁡ ( sin ⁡ ( 1 ) ) ) 5 \begin{aligned} (f^{-1})^{\prime}(x) &= \frac{1}{ f^{\prime}(f^{-1}(x)) } \\ (f^{-1})^{\prime \prime}(x) &= - \frac{1}{ (f^{\prime}(f^{-1}(x)))^2 } \cdot f^{\prime \prime}(f^{-1}(x)) \cdot (f^{-1})^{\prime}(x) \\ (f^{-1})^{\prime \prime}(x) &= - \frac{1}{ (f^{\prime}(f^{-1}(x)))^2 } \cdot f^{\prime \prime}(f^{-1}(x)) \cdot \frac{1}{ f^{\prime}(f^{-1}(x)) } \\ (f^{-1})^{\prime \prime}(x) &= - \frac{ f^{\prime \prime}(f^{-1}(x)) }{ (f^{\prime}(f^{-1}(x)))^3 } \\ (f^{-1})^{\prime \prime}(3) &= - \frac{ f^{\prime \prime}(f^{-1}(3)) }{ (f^{\prime}(f^{-1}(3)))^3 } \\ &= - \frac{ f^{\prime \prime}(0) }{ (f^{\prime}(0))^3 } \\ &= - \frac{ 2 \sin(\sin(1)) \cdot \cos(\sin(1)) \cdot \cos(1) }{ ((\sin(\sin(1)))^2)^3 } \\ &= - \frac{ 2 \sin(\sin(1)) \cdot \cos(\sin(1)) \cdot \cos(1) }{ (\sin(\sin(1)))^6 } \\ &= - \frac{ 2 \cos(\sin(1)) \cdot \cos(1) }{ (\sin(\sin(1)))^5 } \\ \end{aligned}

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