If x = log t and y + 1 = frac{1}{t},then e^{- | vedclass

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

Question

(A) Given $x = \log t$ and $y + 1 = \frac{1}{t}$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = -\f

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given $x = \log t$ and $y + 1 = \frac{1}{t}$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = -\frac{1}{t^{2}}$.Now,find $\frac{dx}{dy}$ using the chain rule:$\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{1/t}{-1/t^{2}} = -t$.Next,find the second derivative $\frac{d^{2}x}{dy^{2}}$:$\frac{d^{2}x}{dy^{2}} = \frac{d}{dy}(-t) = \frac{d}{dt}(-t) \cdot \frac{dt}{dy} = (-1) \cdot \frac{1}{dy/dt} = (-1) \cdot \frac{1}{-1/t^{2}} = t^{2}$.Also,calculate $e^{-x}$:$e^{-x} = e^{-\log t} = e^{\log(t^{-1})} = \frac{1}{t}$.Finally,substitute these values into the expression $e^{-x} \frac{d^{2}x}{dy^{2}} + \frac{dx}{dy}$:$\left(\frac{1}{t}\right)(t^{2}) + (-t) = t - t = 0$.

If $x = \log t$ and $y + 1 = \frac{1}{t}$,then $e^{-x} \frac{d^{2} x}{d y^{2}} + \frac{d x}{d y} = $

Similar Questions

If $x=a(t+\sin t)$ and $y=a(1-\cos t)$,then find $\frac{d^2 y}{d x^2}$.

The slope of the tangent at $(1, 2)$ to the curve $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$ is:

Find the slope of the normal to the curve $x = 1 - a \sin \theta$,$y = b \cos^2 \theta$ at $\theta = \pi / 2$.

If $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$,then at $t = \frac{\pi}{4}$,find $\frac{dy}{dx}$.

Derivative of $\sin ^2 x$ with respect to $e^{\cos x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module