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Math

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PROBLEM

The Differential of a Function.
Find the differential of the given function. Then, evaluate the differential at the indicated values.
If y=xcos(x)y=x \cos (x) then the differential of yy is
dy=d y=\square Note: Type dxd x for the differential of xx
Evaluate the differential of yy at x0=4.71239,dx=0.3x_{0}=4.71239, d x=0.3
dyx=x0dx=0.3=\left.d y\right|_{\substack{x=x_{0} \\ d x=0.3}}=\square Note: Enter your answer accurate to 4 decimal places if it is not an Integrer.

STEP 1

What is this asking?
We need to find a formula for the tiny change in yy when there's a tiny change in xx, given that y=xcos(x)y = x \cos(x).
Then, we need to calculate this tiny change in yy when xx is 4.71239 and the tiny change in xx is 0.3.
Watch out!
Remember the product rule for derivatives!
Also, make sure your calculator is in radians mode when evaluating the cosine function.

STEP 2

1. Find the Differential
2. Evaluate at Given Values

STEP 3

We are given the function y=xcos(x)y = x \cos(x).

STEP 4

To find the differential, we need to find the derivative of yy with respect to xx.
Since yy is a product of two functions, xx and cos(x)\cos(x), we'll use the product rule:
dydx=ddx(x)cos(x)+xddx(cos(x)). \frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos(x) + x \cdot \frac{d}{dx}(\cos(x)).

STEP 5

The derivative of xx with respect to xx is simply 1.
The derivative of cos(x)\cos(x) with respect to xx is sin(x)- \sin(x).
So,
dydx=1cos(x)+x(sin(x))=cos(x)xsin(x). \frac{dy}{dx} = 1 \cdot \cos(x) + x \cdot (-\sin(x)) = \cos(x) - x \sin(x).

STEP 6

The differential dydy is given by:
dy=dydxdx=(cos(x)xsin(x))dx. dy = \frac{dy}{dx} dx = (\cos(x) - x \sin(x)) dx.

STEP 7

We are given x0=4.71239x_0 = 4.71239 and dx=0.3dx = 0.3.
Let's plug these values into our differential:
dy=(cos(4.71239)4.71239sin(4.71239))0.3. dy = (\cos(4.71239) - 4.71239 \cdot \sin(4.71239)) \cdot 0.3.

STEP 8

Using a calculator in radians mode:
cos(4.71239)0\cos(4.71239) \approx 0, and sin(4.71239)1\sin(4.71239) \approx -1.

STEP 9

Substituting these values back into the equation:
dy(04.71239(1))0.3=(4.71239)0.31.413717. dy \approx (0 - 4.71239 \cdot (-1)) \cdot 0.3 = (4.71239) \cdot 0.3 \approx 1.413717. Rounding to 4 decimal places, we get dy1.4137dy \approx 1.4137.

SOLUTION

The differential of yy is dy=(cos(x)xsin(x))dxdy = (\cos(x) - x \sin(x)) dx.
At x0=4.71239x_0 = 4.71239 and dx=0.3dx = 0.3, the differential is approximately 1.41371.4137.

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