When is tangent line horizontal?

When is the tangent line of the following function horizontal?
$$y =\frac{\sin x}{e^x}$$

What steps or how can I draw a graph to figure this out or prove this?

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Take the derivative of $y$ with respect to $x$ and set it equal to zero. Find all those $x$'s that satisfy $y'(x)=0$. Recall from class or from your book that setting $y'(x)=0$ is equivalent to having no slope at this point $x$. – math-visitor Jun 25 '12 at 1:08

Apply the quotient rule to find $$y' = \frac{e^x \cos x - e^x \sin x}{e^{2x}}$$

We want $y'=0$, which means we want the numerator to equal zero:

$$0 = e^x \cos x - e^x \sin x$$ $$0 = \cos x - \sin x$$ $$\sin x = \cos x$$ Divide by $\cos x$: $$\tan x = 1$$

This happens at $x = \frac{\pi}{4}$, and since tangent has period $\pi$, it will happen also when we add an integer multiple of $\pi$. Thus, we have a horizontal tangent line at $$x = \frac{\pi}{4} + n\pi, n \in \mathbb{Z}$$

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Okay see what i missed was doing the derivative i totally didnt get it – soniccool Jun 25 '12 at 1:17
How does the $e^2x$ get cancelled out just curious because it = 1? – soniccool Jun 25 '12 at 1:19
We want a fraction to be equal to zero, and the only way to make that happen is if the numerator is zero while the denominator is non-zero. Since $e^{2x}$ is never zero, I did not need to worry about it at all. – Nicholas Kirchner Jun 25 '12 at 1:21
Got it so basically the horizontal tangent line is at tanx? Or $π /4$ Because how do we get $π /4$ out of tanx =1? – soniccool Jun 25 '12 at 1:23
That's something folks are told to memorize in trigonometry. Maybe the easiest way to see it is to recall that, in a right triangle, the tangent of an angle is opposite/adjacent. Since we want $\tan x = 1$, that means opposite = adjacent, so we've gotta be talking about a 45 degree angle. – Nicholas Kirchner Jun 25 '12 at 1:26

If we have a function $y = f(x)$, then the derivative of $y$ with respect to $x$ at the point $x_0$ gives us the slope of the tangent at $x_0$. If you want the tangent to be horizontal, this means that the slope of the tangent must be $0$. Hence, we want to find $x_0$ such that $\left. \dfrac{dy}{dx} \right \rvert_{x_0} = 0$.

In your case, we have $y = \sin(x) \exp(-x)$. Hence, for the tangent to be horizontal at $x_0$, we need $$\left. \dfrac{dy}{dx} \right \rvert_{x_0} = -\sin(x_0) \exp(-x_0) + \cos(x_0) \exp(-x_0) = 0$$ Note that $\exp(-x) >0, \forall x \in \mathbb{R}$. Hence, we get that $$- \sin(x_0) + \cos(x_0) = 0 \implies \tan(x_0) = 1 \implies x_0 = n\pi + \dfrac{\pi}4, \, \forall n \in \mathbb{Z}$$ Hence, at all these points i.e. at $x_0 = n\pi + \dfrac{\pi}4, \, \forall n \in \mathbb{Z}$, we have the tangent being horizontal.

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