Find the $\Delta y$ of $f(x)={1 \over x^2}$; $x=2; \Delta x = 0.01$

Question

Find the $\Delta y$ of $f(x)={1 \over x^2}$; $x=2; \Delta x = 0.01$

when $\Delta y = f(x+ \Delta x) - f(x)$

So here's what I did:

$$\Delta y = f(x+ \Delta x) - f(x) \\ \Delta y = {1 \over (x+ \Delta x)^2} - {1 \over x^2}\\ = {1 \over x^2+2x\Delta x +\Delta x^2} - {1 \over x^2}\\ ={1 \over 2x\Delta x + \Delta x^2}$$ Now substituting x=2 and $\Delta x$=0.01

$${1\over 2(2)(0.01) + (0.01)^2}\\ ={1\over 0.04+0.0001}\\ ={1 \over 0.0401}\\ $$ I'm pretty sure there is something wrong I did. It says that the answer is supposed to be $$-0.00248$$

Answer 1

Everything up to your 3rd step is fine. You could actually just sub in your values here but I'll show you how to properly manipulate fractions by getting them in terms of the LCD. So: $${1 \over x^2+2x\Delta x +\Delta x^2} - {1 \over x^2}=\frac{x^2}{x^2(x^2+2x\Delta x +\Delta x^2)}-\frac{x^2+2x\Delta x +\Delta x^2}{x^2(x^2+2x\Delta x +\Delta x^2)}$$ $$=\frac{-(2x\Delta x+\Delta x^2)}{x^2(x^2+2x\Delta x +\Delta x^2)}$$ Now we substitute in our values: $$\Delta y=\frac{-(4(0.01)+0.01^2)}{4(4+4(0.01)+0.01^2)}=-0.002488$$

Answer 2

$\dfrac{1}{x^2+2x\triangle x +\triangle x^2} - \dfrac{1}{x^2} \neq \dfrac{1}{x^2 +2x\triangle x}$