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

$y=e^{a \sin ^{-1} x} \Rightarrow (1-x^2) y_{n+2}-(2 n+1) x y_{n+1}$ is equal to

$\frac{d^2}{dx^2}(2\cos x \cos 3x) = $

If $f(x) = \sin x + \cos x$, then $f\left(\frac{\pi}{4}\right) f^{(iv)}\left(\frac{\pi}{4}\right)$ is equal to

If ${I_n} = \frac{d^n}{dx^n}(x^n \log x)$,then ${I_n} - n{I_{n - 1}} = $

If $f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module