Question

Find the equation of tangent at $x=0$ on the curve $y=x{e}^x$

• A. y+x = 0
• B. y-x = 0
• C. y + ln 2 x = 0
• D. y-ln2 x = 0

Answer 1

The correct option is B

y-x = 0

To find the equation of the tangent at a given point, we need to know the slope of the tangent. If we know the slope, then we can use the given point $(x_1,y_1)$and write the equation of the tangent.

It will be of the form

$y-y_1=f^{\prime }\left(x_1\right)(x-x_1)$

We have to find the tangent at the point where the x coordinate is zero. We will first find the y coordinate at this point before we find the slope.

Substituting $x=0$ in $x{e}^x$ , we get $y=0$

$\Rightarrow (x_1,y_1)=(0,0)$

$f^{\prime }\left(x\right)=x{e}^x+e^x$

$\Rightarrow f^{\prime }\left(0\right)=1$

$\Rightarrow$ Equation of tangent is $y-0=1(x-0)$

$\Rightarrow y=x$