Don't use calculus

let the sides be x after 10% increase it becomes x+10%x=110/100x older area was x^2 new area is 121/100x^2 difference is 21/100x 21/100x is 21%of x so ans =21%

Original side of a square is $x \Rightarrow$ area is $x^2$

Increased $10\%$ side of a square is $1.1x \Rightarrow$ area is $1.21x^2$

$\frac{1.21x^2}{x^2}=1.21 \Rightarrow$ area increased in $\boxed{21 \% }$

x . x = x²

---

(x + 10% . x) . ( x + 10% x)

(110% . x) . (110% . x)

(11x/10) . (11x / 10)

121x² / 100

1,21 x²

---

1,21 x² - x²

0,21

21 / 100

21%

Let $x$ be the side length of the original square, then the side length of the bigger square is $1.1x$ . The area of the original square is $x^2$ and the area of the bigger square is $(1.1x)^2=1.21x^2$ . So the percent increased in area is $\left(\dfrac{1.21-1}{1}\right) \times 100\% = \boxed{21\%}$ .

Let $a$ and $A$ be the area of the smaller and bigger square respectively. Since they are similar, we have

$\dfrac{a}{A}=\dfrac{1^2}{1.1^2}$

$\dfrac{a}{A}=\dfrac{1}{1.21}$

$A=1.21a$

Therefore,

$\%~increased=(1.21-1)(100)=21\%$

let the side be 10cm 10x10=100cm & 10% increased=11 so the 10%increased side is 11cm 11x11=121cm if 100cm =100% 121cm =121% so,the % increased in area =21%

Differentiation can be used only for small change(<5% say)