The equation of the tangent to the curve y=4 | vedclass

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Question

(B) Given curve is $y=4 x e^{x}$.First,find the derivative $\frac{dy}{dx}$ using the product rule:$\frac{dy}{dx} = 4e^{x} + 4x e^{x} = 4e^{x}(1+x)$.No

Vedclass · Accepted Answer

$y=-\frac{4}{e}$

Vedclass · Answer

(B) Given curve is $y=4 x e^{x}$.First,find the derivative $\frac{dy}{dx}$ using the product rule:$\frac{dy}{dx} = 4e^{x} + 4x e^{x} = 4e^{x}(1+x)$.Now,evaluate the slope at the point $\left(-1, -\frac{4}{e}\right)$:$\left(\frac{dy}{dx}\right)_{(-1, -4/e)} = 4e^{-1}(1 + (-1)) = 4e^{-1}(0) = 0$.Since the slope is $0$,the tangent is a horizontal line.The equation of the tangent line is given by $y - y_1 = m(x - x_1)$:$y - (-\frac{4}{e}) = 0(x - (-1))$.$y + \frac{4}{e} = 0$.Therefore,$y = -\frac{4}{e}$.

The equation of the tangent to the curve $y=4 x e^{x}$ at $\left(-1, -\frac{4}{e}\right)$ is

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