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

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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

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