Question Find $\Delta y$ , $dy=f'(x)dx$ , and $dy$ for $y=2x^3$ , $x=4$ , and $\Delta x=0.08$ . a) $\Delta y=\boxed{1.6384}$ b) $dy=f'(x)dx=\boxed{96}dx$ c) $dy=\boxed{7.68}$

Studdy Solution

STEP 1

1. The function $f(x) = 2x^3$ is differentiable, and we can find its derivative $f'(x)$ .
2. $\Delta y$ represents the actual change in $y$ when $x$ changes by $\Delta x$ .
3. $dy = f'(x)dx$ represents the differential of $y$ , which approximates the change in $y$ for a small change in $x$ .
4. Rounding should be done after calculations are completed for each part.

STEP 2

1. Calculate $\Delta y$ for the given $x$ and $\Delta x$ values.
2. Find the derivative $f'(x)$ of the function $f(x)$ .
3. Calculate $dy$ for the given $x$ and $\Delta x$ values.

STEP 3

Calculate the value of $f(x)$ at $x = 4$ .
$f(4) = 2 \cdot 4^3$

STEP 4

Calculate the value of $f(x)$ at $x = 4 + \Delta x$ where $\Delta x = 0.08$ .
$f(4 + \Delta x) = f(4 + 0.08) = 2 \cdot (4 + 0.08)^3$

STEP 5

Calculate $\Delta y = f(4 + \Delta x) - f(4)$ .
$\Delta y = 2 \cdot (4 + 0.08)^3 - 2 \cdot 4^3$

STEP 6

Simplify the expression for $\Delta y$ and round to four decimal places as needed.

STEP 7

Find the derivative $f'(x)$ of the function $f(x) = 2x^3$ .
$f'(x) = \frac{d}{dx}(2x^3)$

STEP 8

Use the power rule to differentiate $2x^3$ .
$f'(x) = 2 \cdot 3x^{3-1}$

STEP 9

Simplify the expression for $f'(x)$ .
$f'(x) = 6x^2$

STEP 10

Calculate $dy = f'(x)dx$ at $x = 4$ and $dx = \Delta x = 0.08$ .
$dy = f'(4) \cdot 0.08$

STEP 11

Substitute the value of $f'(4)$ into the expression for $dy$ .
$dy = 6 \cdot 4^2 \cdot 0.08$

STEP 12

Simplify the expression for $dy$ and round to two decimal places as needed.
Now, let's carry out the calculations for each step.
STEP_1: Calculate the value of $f(4)$ .
$f(4) = 2 \cdot 4^3 = 2 \cdot 64 = 128$
STEP_2: Calculate the value of $f(4 + \Delta x)$ .
$f(4 + 0.08) = 2 \cdot (4 + 0.08)^3$
STEP_3: Calculate $\Delta y$ .
$\Delta y = 2 \cdot (4 + 0.08)^3 - 128$
STEP_4: Simplify $\Delta y$ and round to four decimal places.
$\Delta y = 2 \cdot (4.08)^3 - 128 \approx 2 \cdot 68.14 - 128 \approx 136.29 - 128 = 8.29$
STEP_5: Find the derivative $f'(x)$ .
$f'(x) = \frac{d}{dx}(2x^3)$
STEP_6: Use the power rule to differentiate $2x^3$ .
$f'(x) = 2 \cdot 3x^{3-1}$
STEP_7: Simplify the expression for $f'(x)$ .
$f'(x) = 6x^2$
STEP_8: Calculate $dy$ at $x = 4$ and $dx = 0.08$ .
$dy = f'(4) \cdot 0.08$
STEP_9: Substitute the value of $f'(4)$ into the expression for $dy$ .
$dy = 6 \cdot 4^2 \cdot 0.08 = 6 \cdot 16 \cdot 0.08$
STEP_10: Simplify the expression for $dy$ and round to two decimal places.
$dy = 6 \cdot 16 \cdot 0.08 = 96 \cdot 0.08 = 7.68$
The solutions are: a) $\Delta y \approx 8.29$ (rounded to four decimal places) b) $dy = f'(x) dx = 6x^2 dx$ c) $dy \approx 7.68$ (rounded to two decimal places)

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