If left(frac{dy}{dx}right) = frac{1}{left(fra | vedclass

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

Question

(A) Given that, $\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1} \dots (i)$Differentiating both sides with respect to $x$ using the chain rule:$\frac{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given that, $\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1} \dots (i)$Differentiating both sides with respect to $x$ using the chain rule:$\frac{d^2y}{dx^2} = -\left(\frac{dx}{dy}\right)^{-2} \cdot \frac{d}{dx}\left(\frac{dx}{dy}\right)$Using the chain rule $\frac{d}{dx} = \frac{dy}{dx} \cdot \frac{d}{dy}$:$\frac{d^2y}{dx^2} = -\left(\frac{dx}{dy}\right)^{-2} \cdot \left(\frac{d^2x}{dy^2} \cdot \frac{dy}{dx}\right)$Since $\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1}$, we substitute this into the equation:$\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^2 \cdot \frac{d^2x}{dy^2} \cdot \frac{dy}{dx}$$\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^3 \cdot \frac{d^2x}{dy^2}$Rearranging the terms:$\frac{d^2x}{dy^2} \left(\frac{dy}{dx}\right)^3 + \frac{d^2y}{dx^2} = 0$Comparing this with the given expression $\frac{d^2x}{dy^2} \left(\frac{dy}{dx}\right)^3 + \frac{d^2y}{dx^2} = k$, we get $k = 0$.Now, evaluating the expression $e^{k f(x)} - k f(x)$:$e^{0 \cdot f(x)} - 0 \cdot f(x) = e^0 - 0 = 1 - 0 = 1$.

If $\left(\frac{dy}{dx}\right) = \frac{1}{\left(\frac{dx}{dy}\right)}$ and $\frac{d^2x}{dy^2}\left(\frac{dy}{dx}\right)^3 + \frac{d^2y}{dx^2} = k$, then $e^{k f(x)} - k f(x) =$

Similar Questions

If $f(x) = \begin{cases} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}$,then

If $e^{y}(x+1)=1,$ show that $\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$.

For $y=7 \sin x+5 \cos x$,if $\frac{d^2 y}{d x^2}-m y=0$,then $m=$ . . . . . .

If $\frac{d^n y}{d x^n}=y_n$ and $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$,then $4 x y_2+2 y_1=$

If $y = \sin^{-1} x$,then $(1 - x^2) y_2 - x y_1 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module