detailed solution

Correct option is @

Let $x=2,x+\mathrm{\Delta }x=1.99.$ Then, $\mathrm{\Delta }x=1.99-2=-0.01.$

Let $dx=\mathrm{\Delta }x=-0.01$

We have,

$y=x^4-10\Rightarrow \frac{dy}{dx}=4x^3\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=2}=4(2{)}^3=32$

Now, $dy=\frac{dy}{dx}dx$

$\Rightarrow dy=32(-0.01)=-0.32$

$\Rightarrow \mathrm{\Delta }y=-0.32$ approximately $[\because \mathrm{\Delta }y\cong dy]$

So, approximate change in $y=-0.32.$

detailed solution

Correct answer is -0.32

Let $x=2,x+\mathrm{\Delta }x=1.99.$ Then, $\mathrm{\Delta }x=1.99-2=-0.01.$

Let $dx=\mathrm{\Delta }x=-0.01$

We have,

$y=x^4-10\Rightarrow \frac{dy}{dx}=4x^3\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=2}=4(2{)}^3=32$

Now, $dy=\frac{dy}{dx}dx$

$\Rightarrow dy=32(-0.01)=-0.32$

$\Rightarrow \mathrm{\Delta }y=-0.32$ approximately $[\because \mathrm{\Delta }y\cong dy]$

So, approximate change in $y=-0.32.$