1) Is this exactly so far? Wolfram doesn"t seem to process this correctly.

You are watching: An equation of the line tangent to the graph of y=x+cosx at the point (0 1) is

2) just how would I increase this to get it in $y$-intercept form? I recognize I can plug $pi$ into my derivative, yet i"m not sure what to make of plugging $x$ right into it. Ns feel prefer it wouldn"t it is in correct, or it would certainly be correct but it would certainly be so verbose the data would certainly be unusable.

Your derivative, i m sorry we need for slope, is close, however $y"$ should be $$y" =underbrace(1)_frac ddx(x)cdot(cos x) + (x)underbrace( -sin x)_frac ddx( cos x)= cos x -xsin x$$ Now, because that slope itself, us evaluate $y"(pi) = cos (pi) - pisin(pi) = -1 - 0 = -1$.

That provides you the equation the the line: $$y+pi = -(x -pi)$$

To get the slope-intercept form, simply distribute the an unfavorable on the right, and also subtract $pi$ from each side to isolate $y$: $$y + pi = -x+ pi iff y = -x$$

Your gradient of the tangent line is no correct.

**Step 1.** distinguish $y=xcos x$. $y"=cos x-xsin x$.

**Step 2.** discover the gradient the the tangent line by putting $x=pi$ right into $y"$; hence $m=-1$.

**Step 3.** create the equation the the heat in the type $y-y_0=m(x-x_0)$; hence $y+pi=-1(x-pi)$ (or $y=-x$)

To uncover the $y$-intercept, you put $x=0$, for this reason $A(0,0)$ is her $y$-intercept.

Recall the the product dominance is

$$dfracddx

Let $f(x) = x$ and $g(x) = cos(x)$. Then, $f"(x) = 1$ and also $g"(x) = -sin(x)$. So the derivative of $y = xcos(x)$ is$$y" = cos(x) - xsin(x)$$

Then, at $x = pi$$$y"(pi) = cos(pi) - pi cdot sin(pi) = -1 - pi cdot 0 = -1$$

Thus, the equation that the tangent heat is$$y + pi = -(x - pi)$$

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