The differentiable functions x and y are related by the following equation: (1)/(y)=cos(x) Also, (dx)/(dt)=-2. Find (dy)/(dt) when x=pi.

Differentiate with respect to t: Differentiate both sides of the equation (1)/(y) = cos(x) with respect to t using implicit differentiation. Since (1)/(y) is y^(-1), we use the chain rule to differentiate y^(-1) with respect to t, which gives us -y^(-2) * (dy)/(dt) on the left side. On the right side, the derivative of cos(x) with respect to t is -sin(x) * (dx)/(dt). So, we have -y^(-2) * (dy)/(dt) = -sin(x) * (dx)/(dt).Substitute given value: Substitute the given value of (dx)/(dt) into the equation. We know that (dx)/(dt) = -2, so we substitute this into the equation to get -y^(-2) * (dy)/(dt) = -sin(x) * (-2).Evaluate sin(x): Evaluate sin(x) when x = pi. Since sin(pi) = 0, we substitute this into the equation to get -y^(-2) * (dy)/(dt) = -0 * (-2).Simplify the equation: Simplify the equation. Since -0 * (-2) is 0, the equation simplifies to -y^(-2) * (dy)/(dt) = 0.Solve for (dy)/(dt): Solve for (dy)/(dt). To find (dy)/(dt), we divide both sides of the equation by -y^(-2). However, since the right side of the equation is 0, (dy)/(dt) must also be 0, regardless of the value of y. So, (dy)/(dt) = 0.

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The differentiable functions
x and
y are related by the following equation:

(1)/(y)=cos(x)
Also,
(dx)/(dt)=-2.
Find
(dy)/(dt) when
x=pi.

The differentiable functions $x$ and $y$ are related by the following equation: $\newline$ $\frac{1}{y}=\cos (x)$ $\newline$ Also, $\frac{d x}{d t}=-2$ . $\newline$ Find $\frac{d y}{d t}$ when $x=\pi$ .

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Algebra 2

Simplify expressions using trigonometric identities

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Q. The differentiable functions

x

and

y

are related by the following equation:

\newline

\frac{1}{y}=\cos (x)

\newline

Also,

\frac{d x}{d t}=-2

\newline

Find

\frac{d y}{d t}

when

x=\pi

Differentiate with respect to $t$ : Differentiate both sides of the equation $\frac{1}{y} = \cos(x)$ with respect to $t$ using implicit differentiation. $\newline$ Since $\frac{1}{y}$ is $y^{-1}$ , we use the chain rule to differentiate $y^{-1}$ with respect to $t$ , which gives us $-y^{-2} \cdot \frac{dy}{dt}$ on the left side. On the right side, the derivative of $\cos(x)$ with respect to $t$ is $\frac{1}{y} = \cos(x)$ $0$ . $\newline$ So, we have $\frac{1}{y} = \cos(x)$ $1$ .
Substitute given value: Substitute the given value of $(dx)/(dt)$ into the equation. $\newline$ We know that $(dx)/(dt) = -2$ , so we substitute this into the equation to get $-y^{-2} * (dy)/(dt) = -\sin(x) * (-2)$ .
Evaluate $\sin(x)$ : Evaluate $\sin(x)$ when $x = \pi$ . $\newline$ Since $\sin(\pi) = 0$ , we substitute this into the equation to get $-y^{-2} \cdot \frac{dy}{dt} = -0 \cdot (-2)$ .
Simplify the equation: Simplify the equation. $\newline$ Since $-0 \times (-2)$ is $0$ , the equation simplifies to $-y^{-2} \times \frac{dy}{dt} = 0$ .
Solve for $(\frac{dy}{dt}):</b> Solve for \$(\frac{dy}{dt})$ . $\newline$ To find $(\frac{dy}{dt})$ , we divide both sides of the equation by $-y^{-2}$ . However, since the right side of the equation is $0$ , $(\frac{dy}{dt})$ must also be $0$ , regardless of the value of $y$ . $\newline$ So, $(\frac{dy}{dt}) = 0$ .

The differentiable functions $x$ and $y$ are related by the following equation: $\newline$ $\frac{1}{y}=\cos (x)$ $\newline$ Also, $\frac{d x}{d t}=-2$ . $\newline$ Find $\frac{d y}{d t}$ when $x=\pi$ .

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The differentiable functions x x x and y y y are related by the following equation:\newline1y=cos⁡(x) \frac{1}{y}=\cos (x) y1​=cos(x)\newlineAlso, dxdt=−2 \frac{d x}{d t}=-2 dtdx​=−2.\newlineFind dydt \frac{d y}{d t} dtdy​ when x=π x=\pi x=π.

The differentiable functions $x$ and $y$ are related by the following equation: $\newline$ $\frac{1}{y}=\cos (x)$ $\newline$ Also, $\frac{d x}{d t}=-2$ . $\newline$ Find $\frac{d y}{d t}$ when $x=\pi$ .