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

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Question

(D) Given the curve equation is $y = 4xe^{x}$.To find the slope of the tangent,we differentiate $y$ with respect to $x$ using the product rule:$\frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the curve equation is $y = 4xe^{x}$.To find the slope of the tangent,we differentiate $y$ with respect to $x$ using the product rule:$\frac{dy}{dx} = 4(e^{x} + xe^{x}) = 4e^{x}(1 + x)$.Now,evaluate the slope at the point $\left(-1, -\frac{4}{e}\right)$:$\left(\frac{dy}{dx}\right)_{x=-1} = 4e^{-1}(1 + (-1)) = 4e^{-1}(0) = 0$.Since the slope of the tangent is $0$,the tangent is a horizontal line parallel to the $X$-axis.The equation of a horizontal line passing through $\left(-1, -\frac{4}{e}\right)$ is $y = y_{1}$,which is $y = -\frac{4}{e}$.

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

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