The differentiable functions x and y are related by the following equation: 3y=cos(x) Also, (dy)/(dt)=5. Find (dx)/(dt) when x=(pi)/(2).

Differentiate with chain rule: First, we need to differentiate both sides of the equation 3y = cos(x) with respect to t, using the chain rule for the right side since x is a function of t.Apply derivative rules: Differentiating the left side with respect to t gives us 3(dy/dt). On the right side, the derivative of cos(x) with respect to x is -sin(x), and then we multiply by (dx/dt) to account for the chain rule.Substitute known values: Setting up the differentiation, we get 3(dy/dt) = -sin(x)(dx/dt).Simplify the equation: We know that (dy/dt) = 5, so we can substitute this value into the equation, which gives us 3(5) = -sin(x)(dx/dt).Find dx/dt at x=(pi)/2: Simplifying the left side, we get 15 = -sin(x)(dx/dt).We need to find (dx/dt) when x = (pi)/(2). At x = (pi)/(2), sin(x) is equal to 1. So, we substitute sin((pi)/(2)) = 1 into the equation.The equation now is 15 = -(1)(dx/dt), which simplifies to 15 = -(dx/dt).To solve for (dx/dt), we divide both sides by -1, which gives us (dx/dt) = -15.

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

3y=cos(x)
Also,
(dy)/(dt)=5.
Find
(dx)/(dt) when
x=(pi)/(2).

The differentiable functions $x$ and $y$ are related by the following equation: $\newline$ $3 y=\cos (x)$ $\newline$ Also, $\frac{d y}{d t}=5$ . $\newline$ Find $\frac{d x}{d t}$ when $x=\frac{\pi}{2}$ .

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Q. The differentiable functions

x

and

y

are related by the following equation:

\newline

3 y=\cos (x)

\newline

Also,

\frac{d y}{d t}=5

\newline

Find

\frac{d x}{d t}

when

x=\frac{\pi}{2}

Differentiate with chain rule: First, we need to differentiate both sides of the equation $3y = \cos(x)$ with respect to $t$ , using the chain rule for the right side since $x$ is a function of $t$ .
Apply derivative rules: Differentiating the left side with respect to $t$ gives us $3\frac{dy}{dt}$ . On the right side, the derivative of $\cos(x)$ with respect to $x$ is $-\sin(x)$ , and then we multiply by $\frac{dx}{dt}$ to account for the chain rule.
Substitute known values: Setting up the differentiation, we get $3\frac{dy}{dt} = -\sin(x)\frac{dx}{dt}$ .
Simplify the equation: We know that $(\frac{dy}{dt}) = 5$ , so we can substitute this value into the equation, which gives us $3(5) = -\sin(x)(\frac{dx}{dt})$ .
Find $\frac{dx}{dt}$ at $x=\frac{\pi}{2}$ : Simplifying the left side, we get $15 = -\sin(x)\left(\frac{dx}{dt}\right)$ .
Find $\frac{dx}{dt}$ at $x=\frac{\pi}{2}$ : Simplifying the left side, we get $15 = -\sin(x)\left(\frac{dx}{dt}\right)$ .We need to find $\frac{dx}{dt}$ when $x = \frac{\pi}{2}$ . At $x = \frac{\pi}{2}$ , $\sin(x)$ is equal to $1$ . So, we substitute $\sin\left(\frac{\pi}{2}\right) = 1$ into the equation.
Find $\frac{dx}{dt}$ at $x=\frac{\pi}{2}$ : Simplifying the left side, we get $15 = -\sin(x)\left(\frac{dx}{dt}\right)$ .We need to find $\left(\frac{dx}{dt}\right)$ when $x = \frac{\pi}{2}$ . At $x = \frac{\pi}{2}$ , $\sin(x)$ is equal to $1$ . So, we substitute $\sin\left(\frac{\pi}{2}\right) = 1$ into the equation.The equation now is $15 = -(1)\left(\frac{dx}{dt}\right)$ , which simplifies to $x=\frac{\pi}{2}$ $0$ .
Find $\frac{dx}{dt}$ at $x=\frac{\pi}{2}$ : Simplifying the left side, we get $15 = -\sin(x)\left(\frac{dx}{dt}\right)$ .We need to find $\left(\frac{dx}{dt}\right)$ when $x = \frac{\pi}{2}$ . At $x = \frac{\pi}{2}$ , $\sin(x)$ is equal to $1$ . So, we substitute $\sin\left(\frac{\pi}{2}\right) = 1$ into the equation.The equation now is $15 = -(1)\left(\frac{dx}{dt}\right)$ , which simplifies to $x=\frac{\pi}{2}$ $0$ .To solve for $\left(\frac{dx}{dt}\right)$ , we divide both sides by $x=\frac{\pi}{2}$ $2$ , which gives us $x=\frac{\pi}{2}$ $3$ .