What is the tangent of f(x)=5xe^x,\at x=0 ?

The tangent of f(x)=5xe^x,\at x=0 is y=5x

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tangent of f(x)=5xe^x,\at x=0

tangent of

f (x) = 5 x e^{x}, at x = 0

y = 5 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = 5 x e^{x} : \frac{df ( x )}{d x} = 5 (e^{x} + e^{x} x)

Compute the slope of

f (x) = 5 x e^{x}

x = 0 : m = 5

Find the line with slope m=

5

and passing through

(0, 0) : y = 5 x

y = 5 x

\frac{d}{d x} (\frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}})

sin (\frac{1}{x ^{2} + 1})

\frac{sin ( x ) - x}{x ^{2}}

(- 2 x y 2 - 1 y) d x + (- 2 x 2 y + 1 x) d y = 0

\frac{d}{d x} (e^{2 x} (2 x - 1))

What is the tangent of f(x)=5xe^x,\at x=0 ?
The tangent of f(x)=5xe^x,\at x=0 is y=5x