The tangent to the curve y = xe^{x^2} passing | vedclass

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

Question

(B) Given the curve $y = xe^{x^2}$.First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = x \cdot e^{x^2}

Vedclass · Accepted Answer

$(\frac{4}{3}, 2e)$

Vedclass · Answer

(B) Given the curve $y = xe^{x^2}$.First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = x \cdot e^{x^2} \cdot (2x) + e^{x^2} \cdot 1 = e^{x^2}(2x^2 + 1)$.At the point $(1, e)$,the slope $m$ is:$m = e^{1^2}(2(1)^2 + 1) = e(2 + 1) = 3e$.The equation of the tangent line at $(1, e)$ is given by $y - y_1 = m(x - x_1)$:$y - e = 3e(x - 1)$$y - e = 3ex - 3e$$y = 3ex - 2e$.Now,we check which point satisfies this equation $y = 3ex - 2e$:For option $B$ $(\frac{4}{3}, 2e)$:$y = 3e(\frac{4}{3}) - 2e = 4e - 2e = 2e$.Since the point $(\frac{4}{3}, 2e)$ satisfies the equation,the tangent passes through this point.

The tangent to the curve $y = xe^{x^2}$ passing through the point $(1, e)$ also passes through the point

Similar Questions

If $(a^2-1) x+a y+(3-a)=0$ is a normal to the curve $x y=1$,then the interval in which '$a$' lies is

The equation of the tangent to the curve $6y = 7 - x^3$ at the point $(1, 1)$ is:

The equation of a normal to the curve $\sin y = x \sin \left( \frac{\pi}{3} + y \right)$ at $x = 0$ is:

If the equation of the normal to the curve $x^3 - y^2 = 0$ at the point $(m^2, -m^3)$ is $y = 3mx - 4m^3$,then $m^2 = \dots\dots$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module