Ify=4 $$ x+3 x^{2} $$, find the approximate change in $$ y $$ when $$ x $$ changes from 9 to 8.98. $$ y(x)=4 \sqrt{x}+3 x^{2} $$ $$ y^{2} $$ fion

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We are given 
$$
y(x)=4\sqrt{x}+3x^{2}.
$$
We want to approximate the change in $$ y $$ when $$ x $$ changes from $$ 9 $$ to $$ 8.98 $$. Using differentials, the approximate change $$ \Delta y $$ is given by

$$
\Delta y\approx y'(x)\,\Delta x,
$$

where $$ \Delta x =-0.02 $$ (since $$ 8.98-9=-0.02 $$).

**Step 1. Find the derivative $$ y'(x) $$.**

Differentiate each term of $$ y(x) $$:

1. For $$ 4\sqrt{x} $$:
   $$
   \frac{d}{dx}(4\sqrt{x}) = 4\cdot\frac{1}{2\sqrt{x}}=\frac{2}{\sqrt{x}}.
   $$

2. For $$ 3x^{2} $$:
   $$
   \frac{d}{dx}(3x^{2}) = 6x.
   $$

Thus, the derivative is

$$
y'(x)=\frac{2}{\sqrt{x}}+6x.
$$

**Step 2. Evaluate the derivative at $$ x=9 $$.**

Since 
$$
\sqrt{9}=3,
$$
we have

$$
y'(9)=\frac{2}{3}+6(9)=\frac{2}{3}+54.
$$

Combine the terms:
$$
y'(9)=\frac{2+162}{3}=\frac{164}{3}.
$$

**Step 3. Calculate the approximate change in $$ y $$.**

Using 
$$
\Delta x=-0.02,
$$
we get

$$
\Delta y\approx y'(9)\,\Delta x=\frac{164}{3}\times(-0.02).
$$

Performing the multiplication:
$$
\Delta y\approx -\frac{164\times0.02}{3}=-\frac{3.28}{3}\approx -1.093.
$$

Thus, the approximate change in $$ y $$ is

$$
\boxed{\Delta y\approx -1.09.}
$$
The approximate change in $$ y $$ when $$ x $$ changes from 9 to 8.98 is about -1.09.

Ify=4 \( x+3 x^{2} \), find the approximate change in \( y \) when \( x \) changes from 9 to 8.98. \( y(x)=4 \sqrt{x}+3 x^{2} \) \( y^{2} \) fion

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