How would you solve to find the equation of the tangent line to the curve $f or y = (1 + x) cos x$ $for y = (1+x) cos x$ at the given point (0,1)?

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Answer 1 · 2018-06-30T03:58:19

$y = x + 1$ $y=x+1$

$y = (1 + x) cos x$ $y=(1+x)cosx$

First let's verify that $(0, 1)$ $(0,1)$ is a point on $y$ $y$

$y (0) = (1 + 0) cos (0) = 1 \times 1 = 1 \to y (0) = 1$ $y(0)= (1+0)cos(0) = 1xx1=1 -> y(0) =1$

$\frac{d y}{d x} = (1 + x) \cdot (- sin x) + 1 \cdot cos x$ $dy/dx= (1+x)*(-sinx) + 1*cosx$

$= cos x - (1 + x) sin x$ $= cosx-(1+x)sinx$

$\frac{d y}{d x}$ $dy/dx$ at $x = 0$ $x=0$ will give us the slope of $y$ $y$ at $(0, 1)$ $(0,1)$

Call this slope $m$ $m$

$\to m = cos (0) - (1 - 0) sin (0) = 1 - 1 \times 0 = 1$ $-> m = cos(0) - (1-0)sin(0) = 1-1xx0 =1$

Now, the tangent to $y$ $y$ at $(0, 1)$ $(0,1)$ will have the equation:

$y = m x + c$ $y=mx+c$ where $m = 1$ $m=1$ (from above)

Since, $(0, 1)$ $(0,1)$ is a point on this line

$:.1=1xx0+c -> c=1$

Hence, our tangent has the equation: $y=x+1$

We can see the graph of $y$ (Red) and our tangent (Blue) in the graphic below.

![enter image source here]
( useruploads.socratic.org )

How would you solve to find the equation of the tangent line to the curve for y = (1+x) cos xfory=(1+x)cosxfor y = (1+x) cos x at the given point (0,1)?