What is the tangent of f(x)=xe^x,\at x=1 ?

The tangent of f(x)=xe^x,\at x=1 is y=2ex-e

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tangent of f(x)=xe^x,\at x=1

tangent of

f (x) = x e^{x}, at x = 1

y = 2 e x - e

Solution steps

Find the tangent point:

(1, e)

Find the slope of

f (x) = x e^{x} : \frac{df ( x )}{d x} = e^{x} + e^{x} x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 e

Find the line with slope m=

2 e

and passing through

(1, e) : y = 2 e x - e

y = 2 e x - e

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x \to 0 + lim (\frac{∣ x - 4 ∣}{x - 4})

\int (3 x^{2} - 8 x + 3) d x

\int x^{2} 7 x - 5 d x

What is the tangent of f(x)=xe^x,\at x=1 ?
The tangent of f(x)=xe^x,\at x=1 is y=2ex-e