If y=f(x) is a thrice differentiable function | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given,$y=f(x)$ is a differentiable and bijective function. We know that $\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} = \left(\frac{d y}{d x}\right

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given,$y=f(x)$ is a differentiable and bijective function. We know that $\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} = \left(\frac{d y}{d x}\right)^{-1}$. Differentiating both sides with respect to $y$: $\frac{d^2 x}{d y^2} = \frac{d}{d y} \left[ \left(\frac{d y}{d x}\right)^{-1} \right]$. Using the chain rule: $\frac{d^2 x}{d y^2} = \frac{d}{d x} \left[ \left(\frac{d y}{d x}\right)^{-1} \right] \cdot \frac{d x}{d y}$. $\frac{d^2 x}{d y^2} = -\left(\frac{d y}{d x}\right)^{-2} \cdot \frac{d^2 y}{d x^2} \cdot \frac{1}{\frac{d y}{d x}}$. $\frac{d^2 x}{d y^2} = -\frac{\frac{d^2 y}{d x^2}}{\left(\frac{d y}{d x}\right)^3}$. Multiplying both sides by $\left(\frac{d y}{d x}\right)^3$: $\frac{d^2 x}{d y^2} \cdot \left(\frac{d y}{d x}\right)^3 = -\frac{d^2 y}{d x^2}$. Therefore,$\frac{d^2 x}{d y^2} \left(\frac{d y}{d x}\right)^3 + \frac{d^2 y}{d x^2} = 0$.

If $y=f(x)$ is a thrice differentiable function and a bijection,then $\frac{d^2 x}{d y^2}\left(\frac{d y}{d x}\right)^3+\frac{d^2 y}{d x^2}=$

Similar Questions

If $y=A e^{m x}+B e^{n x}$,show that $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$.

If $y = a \sin x + (5 + 2x) \cos x$,then $y'' + y =$

If $y=e^{4x} \cos 5x$,then $\frac{d^{2}y}{dx^{2}}$ at $x=0$ is

If $f(x) = (x^2 - 1)^7$,then $f^{(14)}(x)$ is equal to

$y = a \cos x + (b + 2x) \sin x \Rightarrow y^{\prime \prime} + y = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module