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

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

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

tangent of

(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

9 x - x^{2}, 2 x

x (x^{2} + 9)

\frac{\partial}{\partial x} (sin (x e^{x^{2} y}))

\frac{\partial}{\partial x} (ln (x^{4} + y^{2}) - z)

2 sin (x) + sin^{2} (x)

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