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

The tangent of f(x)=e^x-xe^x,\at x=0 is y=1

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

tangent of

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

y = 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 1) : y = 1

y = 1

x \to - 11 lim (x^{2} + 11)

x \to 0 lim (arctan (e^{x}))

\int_{2}^{3} \frac{1}{( x + 1 ) ln ( x + 1 )} d x

\int \frac{csc ^{2} ( x )}{cot ^{4} ( x )} d x

f (x) = 10 x^{3} - 20 x + \frac{1}{5}

What is the tangent of f(x)=e^x-xe^x,\at x=0 ?
The tangent of f(x)=e^x-xe^x,\at x=0 is y=1