What is the tangent of f(x)=(1+x)cos(x),\at x=0 ?

The tangent of f(x)=(1+x)cos(x),\at x=0 is y=x+1

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tangent of f(x)=(1+x)cos(x),\at x=0

tangent of

f (x) = (1 + x) cos (x), at x = 0

y = x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = (1 + x) cos (x) : \frac{df ( x )}{d x} = cos (x) - sin (x) (1 + x)

Compute the slope of

f (x) = (1 + x) cos (x)

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 1) : y = x + 1

y = x + 1

y = \frac{1}{3} x^{2} - 2 x + 4, (3, 1)

\frac{d}{d x} (cos (tan (4 x)))

\frac{d}{d x} (\frac{x ^{3} - 6 x ^{2} + 5}{x})

n = 0 \sum \infty (\frac{1}{6})^{n}

\frac{\partial}{\partial y} (e^{x y})

What is the tangent of f(x)=(1+x)cos(x),\at x=0 ?
The tangent of f(x)=(1+x)cos(x),\at x=0 is y=x+1