Question

The approximate value of ${\left\{(3.92{)}^2+3(2.1{)}^4\right\}}^{1/6}$ is

• A. 2.466
• B. 3.567
• C. 1.562
• D. 2.577

Answer 1

Answer: (A)

Let $f(x,y)={\left(x^2+3y^4\right)}^{1/6}$
Taking $x=4,\Delta x=-0.08$ and $y=2,\Delta y=0.1$
Differentiating (1) w.r.t. $x$, treating $y$ as constant,
$\therefore \frac{\Delta f}{\Delta x}=\frac{1}{6}{\left(x^2+3y^4\right)}^{-5/6}(2x)$
$=\frac{8}{6}(16+48{)}^{-5/6}=\frac{4}{3}\times 2^{-5}=\frac{1}{24}$
and differentiating (1) w.r.t. $y$ treating $x$ as constant,
$\therefore \frac{\Delta f}{\Delta y}=\frac{1}{6}{\left(x^2+3y^4\right)}^{-5/6}\left(12y^3\right)$
$=\frac{12(8)}{6}(64{)}^{-5/6}=16(2{)}^{-5}=\frac{1}{2}$
$\therefore df=\frac{\Delta f}{\Delta x}\cdot dx+\frac{\Delta f}{\Delta y}dy$
$=\frac{1}{24}\times -0.08+\frac{1}{2}\times 0.1$
$=-\frac{0.01}{3}+\frac{0.1}{2}=0.466$
$\therefore {\left\{(3.92{)}^2+3(2.1{)}^4\right\}}^{1/6}=f(4,2)+df$
$=2+0.466=2.466$