trying to approximately compute $0.93^{2.98}$ without using a calculator

Question

I tried to do that using this method :

let $f(x,y) = x^y$ so now I'm looking for $f(0.93,2.98)$

$\Delta f = f(1,3)-f(0.93,2.98) \approx df = f_xdx+f_ydy$

here $x=1,y=3,dx=-0.07,dy=-0.02$

$f_x=yx^{y-1},f_y=x^y\ln x,f(1,3)=1$

so $$df = 1\cdot(-0.07)+0\cdot(-0.02) = -0.07$$

$$df = -0.07\approx f(1,3)-f(0.93,2.98) \implies 0.93^{2.98} \approx 1.07 $$

but the result I got is a bit far from the real value of $0.93^{2.98}$ ($0.80533$)

my question is : is my reasoning false or is it just a bad approximation ?

Answer 1

The method is fine.

$f_x=yx^{y-1},f_y=x^y\ln x,f(1,3)=1$

You have $f_x(x,y)=yx^{y-1} \implies f_x(1,3)=3 \cdot 1^{2} = 3$; so

$$df = \color{red}{1}\cdot(-0.07)+0\cdot(-0.02) = -0.07$$

should be:

$$df = \color{green}{3}\cdot(-0.07)+0\cdot(-0.02) = \color{blue}{-0.21}$$

which gives:

$$0.93^{2.98} = f(0.93,2.98) \approx f(1,3)\color{blue}{-0.21} = 0.79$$