Is It Just The Reciprocal?

Calculus Level 2

y = f ( x ) , p = d y d x , q = d 2 y d x 2 y = f(x), p = \dfrac{dy}{dx}, q = \dfrac{d^2y}{dx^2}

What is d 2 x d y 2 ? \dfrac{d^2x}{dy^2} \ ?

− p q 3 \frac{-p}{q^3} 1 q \frac{1}{q} p 2 p^2 − q p 3 \frac{-q}{p^3}

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Vishwak Srinivasan by Vishwak Srinivasan , 24, India

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4 solutions

d 2 x d y 2 = d d y ( d x d y ) = d d y ( 1 p ) = − 1 p 2 ˙ d p d y = − 1 p 2 ˙ d p d x ˙ d x d y = − 1 p 2 ˙ d 2 y d x 2 ˙ d x d y = − 1 p 2 ˙ q ˙ 1 p = − q p 3 \begin{aligned} \frac{d^2x}{dy^2} & = \frac{d}{dy} \left( \frac{dx}{dy} \right) = \frac{d}{dy} \left( \frac{1}{p} \right) = -\frac{1}{p^2} \dot{} \frac{dp}{dy} = -\frac{1}{p^2} \dot{} \frac{dp}{dx} \dot{} \frac{dx}{dy} \\ & = -\frac{1}{p^2} \dot{} \frac{d^2y}{dx^2} \dot{} \frac{dx}{dy} = -\frac{1}{p^2} \dot{} q \dot{} \frac{1}{p} = \boxed {- \dfrac {q}{p^3}} \end{aligned}

30 Helpful 4 Interesting 4 Brilliant 2 Confused

Moderator note:

How can one motivate the sequence of steps taken above? Are we going on a blind fishing expedition hoping to land the target, or is there a more directed approach?

Frankly I was motivated by p = d y d x ⇒ d x d y = 1 p p = \frac{dy}{dx} \quad \Rightarrow \frac{dx}{dy} = \frac{1}{p} and d p d x = d 2 y d x 2 \frac{dp}{dx} = \frac{d^2y}{dx^2} . I was quite sure that I can solve it by expressing it in p p .

Chew-Seong Cheong - 5 years, 11 months ago

Sir, thanks for the solution. A minor typo is there. In the first line.

Vishwak Srinivasan - 5 years, 11 months ago

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I don't see any typo.

Chew-Seong Cheong - 5 years, 11 months ago

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Haha. Nice trick sir. There was a '==' in the first line.

Vishwak Srinivasan - 5 years, 11 months ago

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@Vishwak Srinivasan – Yes, there was.

Chew-Seong Cheong - 5 years, 11 months ago

Where is the minus sign come from?

Hafizh Ahsan Permana - 5 years, 11 months ago

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ya I was confused as well, especially with all the weird variables that I ain't used to, but I got it now, they used the chain rule i.e. the derivative of 1/p with respect to y equals the derivative of 1/p with respect to p times the derivative of p with respect to y. switch the p's to x's and treat x as a function of y i.e. x=g(y) and you will get it

Oximas omar - 1 month, 1 week ago

in the third equality, you say -1/p^2 *dp/dy can someone explain where this came from? and why is there a negative?

Oximas omar - 1 month, 1 week ago

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d d y ( 1 p ) = d d y p − 1 = − p − 2 ⋅ d p d y = − 1 p 2 ⋅ d p d y \frac d{dy} \left(\frac 1p\right) = \frac d{dy} p^{-1} = -p^{-2} \cdot \frac {dp}{dy} = - \frac 1{p^2} \cdot \frac {dp}{dy}

Chew-Seong Cheong - 1 month, 1 week ago

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oh okay I get it, it's just the chain rule, thanks

Oximas omar - 1 month, 1 week ago
Fidel R.
Jul 6, 2015

y = f ( x ) y = f(x) d d y y = d d y f ( x ) \frac{d}{dy} y = \frac{d}{dy} f(x) The left side is easy. Use chain rule for the right side. 1 = ( d d x f ( x ) ) ( d x d y ) 1 = (\frac{d}{dx} f(x))(\frac{dx}{dy}) 1 = f ′ ( x ) ( d x d y ) 1 = f'(x)(\frac{dx}{dy}) 1 f ′ ( x ) = d x d y \frac{1}{f'(x)} = \frac{dx}{dy} d x d y = [ f ( x ) ] − 1 \frac{dx}{dy} = [f(x)]^{-1} d d y ( d x d y ) = d d y [ f ′ ( x ) ] − 1 \frac{d}{dy}(\frac{dx}{dy}) = \frac{d}{dy}[f'(x)]^{-1} Use the chain rule on the right side twice. d 2 x d y 2 = ( d d x [ f ′ ( x ) ] − 1 ) ( d x d y ) \frac{d^{2}x}{dy^{2}} = (\frac{d}{dx}[f'(x)]^{-1})(\frac{dx}{dy}) d 2 x d y 2 = ( − [ f ′ ( x ) ] − 2 ) ( f ′ ′ ( x ) ) ( d x d y ) \frac{d^{2}x}{dy^{2}} = (-[f'(x)]^{-2})(f''(x))(\frac{dx}{dy}) d 2 x d y 2 = ( − [ f ′ ( x ) ] − 2 ) ( f ′ ′ ( x ) ) ( [ f ′ ( x ) ] − 1 ) \frac{d^{2}x}{dy^{2}} = (-[f'(x)]^{-2})(f''(x))([f'(x)]^{-1}) d 2 x d y 2 = ( − [ f ′ ( x ) ] − 3 ) ( f ′ ′ ( x ) ) \frac{d^{2}x}{dy^{2}} = (-[f'(x)]^{-3})(f''(x)) d 2 x d y 2 = − f ′ ′ ( x ) [ f ′ ( x ) ] 3 \frac{d^{2}x}{dy^{2}} = - \frac{f''(x)}{[f'(x)]^{3}} Substitute f'(x) with p, and f''(x) with q. d 2 x d y 2 = − q p 3 \boxed{\frac{d^{2}x}{dy^{2}} = -\frac{q}{p^{3}}}

13 Helpful 1 Interesting 3 Brilliant 1 Confused

A little typo on line 6 but understandable

Bostang Palaguna - 7 months, 2 weeks ago
Adarsh Kumar
Jul 3, 2015

We have that, d y d x = p , q = d ( d y d x ) d x w e h a v e t o f i n d t h e v a l u e o f d ( d x d y ) d y d ( 1 p ) d y = a ( s a y ) , 1 q = d x d p ⟹ a q = 1 p d ( 1 p ) d p ⟹ a q = 1 p ( − 1 p 2 ) \dfrac{dy}{dx}=p,q=\dfrac{d(\dfrac{dy}{dx})}{dx}\\we\ have\ to\ find\ the\ value\ of\\ \dfrac{d(\dfrac{dx}{dy})}{dy}\\ \dfrac{d(\dfrac{1}{p})}{dy}=a(say),\dfrac{1}{q}=\dfrac{dx}{dp}\\ \Longrightarrow \dfrac{a}{q}=\dfrac{1}{p} \dfrac{d(\dfrac{1}{p})}{dp}\\ \Longrightarrow \dfrac{a}{q}=\dfrac{1}{p} (-\dfrac{1}{p^2}) .From here we can easily find out the value of a a .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

It is often better to write out all of the steps that were taken, so that the substitutions are clear to follow. IE

d 1 p d y = d 1 p d p × d p d x × d x d y \frac{ d \frac{1}{p} } { dy } = \frac{ d \frac{1}{p} } { dp} \times \frac{dp}{dx} \times \frac{ dx}{dy}

Aditya Kumar
Jul 2, 2015

d x / d y = 1 / p {dx}/{dy} = 1/p

=> d 2 x / d y 2 = − 1 / p 2 ∗ q / p {d{^2}x}/{dy{^2}} = -1/{p{^2}}*q/p

1 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you reach that conclusion? What was the intermediary step?

Calvin Lin Staff - 5 years, 11 months ago

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pure intuition XD

Oximas omar - 1 month, 1 week ago

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