Limits and functions

Calculus Level 3

Given,

y = 1 x 2 + 1 \large y=\frac{1}{x^{2}+1}

Find y ′ y^{\prime} using the definition of the derivative.

x = tan ⁡ α x=\tan \alpha and y = cos ⁡ 2 α y=\cos^2\alpha so that d y d x = d y d α d x d α \dfrac{dy}{dx}=\dfrac{\frac{dy}{d\alpha}}{\frac{dx}{d\alpha}} .

If y ′ y^{\prime} can be expressed as α x x β + 2 x γ + δ \large\frac {\alpha x }{ x^{ \beta }+2x^{ \gamma }+\delta } .

What is the value of α + β + γ + δ \alpha+\beta+\gamma+\delta ?


The answer is 5.

Armain Labeeb by Armain Labeeb , 20, Bangladesh

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1 solution

Armain Labeeb
Jul 10, 2016

L e t f ( x ) = 1 x 2 + 1 T h e n y ′ = f ′ ( x ) = lim ⁡ h → 0 ( f ( x + h ) − f ( x ) h ) = lim ⁡ h → 0 ( 1 ( x + h ) 2 + 1 − 1 x 2 + 1 h ) = lim ⁡ h → 0 ( ( x 2 + 1 ) − ( ( x + h ) 2 + 1 ) ( ( x + h ) 2 + 1 ) ⋅ ( x 2 + 1 ) ⋅ 1 h ) = lim ⁡ h → 0 ( x 2 + 1 − x 2 − 2 x h − h 2 − 1 ( ( x + h ) 2 + 1 ) ⋅ ( x 2 + 1 ) ⋅ h ) = lim ⁡ h → 0 ( h ( − 2 x − h ) h ( ( x + h ) 2 + 1 ) ( x 2 + 1 ) ) = lim ⁡ h → 0 ( ( − 2 x − h ) ( ( x + h ) 2 + 1 ) ( x 2 + 1 ) ) = lim ⁡ h → 0 ( ( − 2 x − 0 ) ( ( x + 0 ) 2 + 1 ) ( x 2 + 1 ) ) = lim ⁡ h → 0 ( − 2 x ( ( x + h ) 2 + 1 ) ( x 2 + 1 ) ) = − 2 x ( x 2 + 1 ) 2 ∴ α x x β + 2 x γ + δ = − 2 x x 4 + 2 x 2 + 1 ⟹ α + β + γ + δ = − 2 + 4 + 2 + 1 = 5 \begin{aligned} Let\quad f\left( x \right) & =\frac { 1 }{ { x }^{ 2 }+1 } \\ Then\quad { y }^{ ' }=f^{ ' }\left( x \right) & =\lim _{ h\to 0 }{ \left( \frac { f\left( x+h \right) -f\left( x \right) }{ h } \right) } \\ & =\lim _{ h\to 0 }{ \left( \frac { \frac { 1 }{ { (x+h) }^{ 2 }+1 } -\frac { 1 }{ { x }^{ 2 }+1 } }{ h } \right) } \\ & =\lim _{ h\to 0 }{ \left( \frac { \left( { x }^{ 2 }+1 \right) -({ (x+h) }^{ 2 }+1) }{ { ((x+h) }^{ 2 }+1)\, \, \cdot \, \, { (x }^{ 2 }+1) } \cdot \frac { 1 }{ h } \right) } \\ & =\lim _{ h\to 0 }{ \left( \frac { { { x }^{ 2 } }+{ 1 }-{ { x }^{ 2 } }-2xh-{ h }^{ 2 }-{ 1 } }{ { ((x+h) }^{ 2 }+1)\, \, \cdot \, \, { (x }^{ 2 }+1)\, \cdot \, h } \right) } \\ & =\lim _{ h\to 0 }{ \left( \frac { { h }(-2x-h) }{ { { h }((x+h) }^{ 2 }+1){ (x }^{ 2 }+1) } \right) } \\ & =\lim _{ h\to 0 }{ \left( \frac { (-2x-h) }{ { ((x+h) }^{ 2 }+1){ (x }^{ 2 }+1) } \right) } \\ & =\lim _{ h\to 0 }{ \left( \frac { (-2x-0) }{ { ((x+0) }^{ 2 }+1){ (x }^{ 2 }+1) } \right) } \\ & =\lim _{ h\to 0 }{ \left( \frac { -2x }{ { ((x+h) }^{ 2 }+1){ (x }^{ 2 }+1) } \right) } \\ & =\frac { -2x }{ { ({ x }^{ 2 }+1) }^{ 2 } } \\ \therefore \frac { \alpha x }{ x^{ \beta }+2x^{ \gamma }+\delta } & =\frac { -2x }{ { { { x }^{ 4 }+2x }^{ 2 }+1 } } \\ \Longrightarrow \alpha +\beta +\gamma +\delta & =-2+4+2+1 \\ & =5 \end{aligned}

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Just for the the sake of variety put x = tan ⁡ α x=\tan \alpha so that y = cos ⁡ 2 α y=\cos^2\alpha . Then d y d x = d y d α d x d α \dfrac{dy}{dx}=\dfrac{\frac{dy}{d\alpha}}{\frac{dx}{d\alpha}} .

Rishabh Jain - 4 years, 11 months ago

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