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

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

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

tangent of

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

y = x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

y = (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

\int \frac{cos ( x )}{sin ^{2} ( x ) sin ( x )} d x

\int \frac{x - 6}{x ^{2} - 18 x + 82} d x

\int t^{2} e^{- 17 t} d t

x \to 3 lim (9 x + 92)

\frac{d}{d x} ((f (x))^{n})

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